STRENGTH  OF  MATERIAL 


AN   ELEMENTARY   STUDY 

PREPARED   FOR  THE  USE  OF  MIDSHIPMEN  AT 
THE   U.  S.  NAVAL  ACADEMY 


BY 

H.  E.  SMITH 
* 

PROFESSOR    OF    MATHEMATICS,    U.    S.    NAVY 


SECOND  EDITION,  REVISED 


NEW  YORK 
JOHN  WILEY  &  SONS,  INC. 

LONDON  :    CHAPMAN   &   HALL,    LIMITED 
1914 


COPYRIGHT,  1908,  1909, 

BY 

H.   E    SMITH,   U.    S.   N. 


Stanhope  Ipresft 

F.    H.   G1LSON     COMPANY 
BOSTON.      USA 


PREFACE. 


THIS  book  has  been  prepared  for  the  use  of  the  Midshipmen 
at  the  U.  S.  Naval  Academy,  and  is  designed  to  cover  a  short 
course  in  the  subject  taken  up  in  the  Department  of  Mathe- 
matics and  Mechanics  preliminary  to  the  work  in  the  Depart- 
ments of  Ordnance  and  Gunnery  and  of  Steam  Engineering 
at  the  Academy. 

In  arranging  the  subject  matter  many  of  the  methods  in- 
troduced by  officers  previously  on  duty  in  the  Department  of 
Mathematics  have  been  employed,  and  the  endeavor  has  been 
to  lead  the  student  to  the  opening  point  for  the  professional 
work  carried  on  by  the  other  Departments. 


iii 

360375 


INTRODUCTION. 


BEFORE  beginning  the  study  of  Strength  of  Material,  let  us 
see  what  has  been  discovered  by  experimenting  with  test 
pieces  of  material  and  note  some  of  the  conclusions  arrived  at 
from  the  results  obtained  in  this  way. 

Experiment  shows  us  that  whenever  a  force  acts  on  a  body 
formed  of  any  substance  the  dimensions  of  that  body  are 
changed.  In  mechanics  all  bodies  were  assumed  rigid  and 
the  results  obtained  under  this  assumption  were  true,  for 
mechanics  taught  us  to  find  the  action  of  one  body  on  another 
or  the  force  transmitted  by  one  body  to  another,  while 
strength  of  material  will  teach  us  the  effect  in  the  body  itself 
of  a  force  acting  upon  it.  Mechanics  showed  us  that  by  means 
of  a  piece  of  material  force  could  be  moved  from  one  point  to 
another,  and  strength  of  material  will  show  us  that  in  trans- 
mitting the  force  the  substance  forming  the  conveyance 
suffers  some  slight  temporary  deformation  if  the  force  is  with- 
in certain  limits;  that  beyond  these  same  limits  the  substance 
suffers  permanent  deformation  and  if  the  force  be  great  enough 
will  be  completely  ruptured. 

The  study  of  strength  of  material  will  include  finding  the 
safe  limits  of  a  force  to  be  transmitted  by  any  particular  piece 
of  material,  finding  the  deformation  caused  by  transmitting 
any  force,  and  finding  the  dimensions  of  a  piece  of  material  in 
order  that  it  may  safely  transmit  any  particular  force. 

With  regard  to  deformation  materials  differ  greatly,  for 
example,  the  force  which  will  double  the  length  of  a  piece 
of  rubber  will  not  apparently  change  the  length  of  a  piece  of 
steel  of  the  same  size,  though  both  of  these  substances  are 
elastic,  and  each  if  stretched  within  limits  will  return  to  its 


VI  INTRODUCTION. 

original  length  when  the  stretching  force  is  removed.  The 
force  which  makes  an  indentation  in  a  piece  of  putty  will 
scarcely  affect  a  piece  of  lead,  but  in  this  case  the  indentation 
made  will  remain  in  both  these  substances  as  they  are  plastic. 
Obviously,  then,  we  must  experiment  with  the  different  ma- 
terials and  find  out  some  of  their  physical  properties  before 
proceeding  with  a  mathematical  investigation. 

The  materials  used  in  building  are  elastic  and  we  determine 
their  physical  properties  by  experimenting  with  small  pieces 
of  them  in  machines  made  for  the  purpose. 

Take  steel, for  example;  small  test  pieces  of  different  shapes 
are  tried  under  different  forces.  A  pull  is  applied  and  we 
find  the  test  pieces  stretch;  if  we  apply  pressure  the  test 
pieces  are  compressed.  Having  applied  all  sorts  of  forces  we 
compare  our  results  and  find  that  the  stretching  or  compres- 
sion is  always  of  the  same  amount  if  the  same  value  of  force 
on  unit  cube  of  the  steel  is  used.  We  also  find  that  up  to  a 
certain  limiting  value  of  the  force  the  material  will  always 
return  to  its  original  shape  when  the  force  is  removed,  but  if 
we  go  beyond  this  value  we  find  the  piece  will  not  return  to 
its  original  shape;  in  other  words  it  is  not  perfectly  elastic 
for  forces  beyond  this  value. 

If  we  go  through  the  whole  list  of  materials  used  in  build- 
ing and  test  each  kind  in  the  same  way  we  will  find  that 
they  will  all  behave  in  a  similar  manner,  the  difference 
will  lie  in  the  amount  of  deformation  for  any  force  and  the 
value  of  the  limiting  force  beyond  which  they  are  not  per- 
fectly elastic. 

Experimenting  further  we  find  the  value  of  this  limiting 
force  for  the  different  materials  and  having  it  we  can  compare 
the  various  substances  as  to  their  usefulness  under  different 
circumstances. 

If  we  continue  to  experiment,  again  and  again  with  a  single 
substance  we  find  that  we  may  apply  as  often  as  we  like  a 
force  less  than  the  limiting  one  we  found,  and  that  the  piece 


INTRODUCTION.  vii 

will  always  return  to  its  original  shape  when  the  force  is 
removed;  but  when  the  force  used  exceeds  the  limiting  one 
found,  if  by  ever  so  little,  and  the  force  is  applied  and  re- 
moved often  enough  the  piece  will  break,  though  the  deforma- 
tion caused  by  the  first  application  of  the  force  was  too  small 
to  be  measured  or  even  noticed.  From  this  fact  we  see  that 
we  must  never  use  a  piece  of  material  which  will  have  to 
sustain  a  force  which  is  in  the  least  greater  than  the  limiting 
value  found  by  experimenting. 

Materials  differ  in  other  ways.  A  piece  of  glass  is  easily 
broken  by  a  light  blow  and  is  therefore  called  brittle  or 
fragile;  wrought  iron  can  be  twisted  and  bent  into  almost  any 
shape  without  rupture  and  therefore  is  called  malleable  or 
ductile. 

Material  which  has  been  melted  and  cast  into  desired  shapes 
cools  quicker  at  some  parts  thaa  it  does  at  others,  thereby 
setting  up  within  it  internal  stresses  which  are  irregularly 
distributed.  Such  castings  can  be  broken  by  a  comparatively 
light  blow  though  they  can  usually  withstand  a  large  pressure. 
These  internal  stresses  can  be  removed  by  a  process  called 
annealing,  which  consists  in  heating  the  body  to  a  red  heat 
and  allowing  it  to  cool  slowly,  thus  allowing  the  particles  an 
opportunity  to  rearrange  themselves. 

Metals  have  the  peculiarity  that  if  overstrained  they  harden 
in  the  vicinity  of  the  overstrain  and  this  hardening  goes  on 
with  time.  Thus  a  bar  which  is  sheared  off  while  cold  will 
finally  become  extremely  hard  and  brittle  near  the  sheared 
end,  as  will  a  plate  in  the  neighborhood  of  cold-punched  rivet 
holes.  To  avoid  this,  bore  the  rivet  holes  and  saw  the  bar,  or, 
if  feasible,  anneal  the  cold-sheared  bar  and  the  plate  near  the 
cold-punched  rivet  holes. 

Now  in  all  practical  cases  we  must  of  course  use  material 
that  will  not  break,  but  in  addition  we  must  have  material 
which  will  not  change  its  dimensions  to  any  considerable 
extent  under  the  applied  load. 


Vlil  INTRODUCTION. 

The  experiments  by  which  the  constants  used  in  this  study 
have  been  determined  were  very  carefully  conducted  and  are 
fthe  results  of  many  independent  efforts  on  the  parts  of  many 
different  scientists.  In  this  work  the  mathematical  investi- 
gation only  will  be  touched  upon,  the  experimental  part 
being  beyond  its  scope. 


CONTENTS. 


PAGE 

INTRODUCTION v 

CHAPTER 

I.   STRESS  AND  STRAIN,  TENSION  AND  COMPRESSION 3 

II.   SHEARING  FORCE 15 

III.  TORSION 26 

IV.  STRESS  DUE  TO  BENDING 34 

V.   COMBINED  STRESSES 43 

VI.   SHEARING  STRESS  IN  BEAMS 52 

VII.    BENDING   MOMENTS,  CURVES  OF  SHEARING  STRESS  AND 

BENDING  MOMENTS 61 

VIII.    SLOPE  AND  DEFLECTION  OF  BEAMS 73 

IX.   SLOPE  AND  DEFLECTION  (Continued) 83 

X.   CONTINUOUS  BEAMS 94 

XI.   COLUMNS  AND  STRUTS 105 

XII.   STRESS  ON  MEMBERS  OF  FRAMES 112 

XIII.  FRAMED  STRUCTURES 123 

XIV.  FRAMED  STRUCTURES  (Continued) 133 

MISCELLANEOUS  PROBLEMS 141 

Reinforced  Concrete  Beams;  Poisson's  Ratio;  Stress  in  Thick 
Cylinders  and  Guns;  Built-up  Guns;  Stress  due  to  Centrifugal 
Force;  Bending  due  to  Centrifugal  Force;  Flat  Plates. 


IX 


STRENGTH   OF   MATERIAL. 


STRENGTH    OF    MATERIAL. 


CHAPTER     I. 
TENSION  AND  COMPRESSION  —  STRESS  AND  STRAIN. 

1.  In  the  study  of  strength  of  material  we  must  consider 
two  ways  of  arranging  the  different  pieces  used:  first,  when 
there  is  to  be  motion  between  the  parts,  and  second,  when  the 
parts  are  to  be  relatively  at  rest.  In  the  first  case,  force  is 
transmitted  from  one  piece  to  another  and  the  combination 
of  pieces  is  called  a  machine,  the  study  of  which  involves  the 
principles  of  dynamics;  the  second  arrangement  is  called  a 
framed  structure,  or  simply  structure,  and  we  must  employ 
the  principles  of  statics  in  its  investigation.  In  either 
arrangement,  any  two  parts  in  contact  have  a  mutual  action 
between  the  touching  surfaces,  and  the  effects  produced  by 
this  action  depend  in  great  measure  upon  the  way  in  which  it 
is  applied.  In  any  case  it  tends  to  change  the  shape  or  dimen- 
sions of  the  parts  involved  and,  if  the  force  is  great  enough,  to 
crush  or  break  them.  So  for  permanence  the  machine  or 
structure  must  be  strong  enough  in  each  part  to  withstand 
any  force  to  which  it  may  be  subjected.  If  then  we  can  find 
the  greatest  force  that  any  particular  piece  of  material  can 
endure  without  breaking  or  suffering  a  permanent  change  in 
shape  we  can  be  sure  of  its  remaining  intact  for  all  force 
within  that  limit.  In  the  first  four  chapters  we  will  apply, 
separately,  all  the  different  forces  to  which  a  piece  of  material 
can  be  subjected  and  as  any  piece  in  our  machine  or  struc- 
ture may  have  more  than  one  of  these  forces  to  sustain  at  any 

3 


4     ..  .STRENGTH  OF  MATERIAL. 

given  instant  we  will,  in  the  fifth  chapter,  show  the  effect  of 
the  combined  action  of  two  or  more  of  them. 

2.  Let  us  first  investigate  the  effect  on  a  piece  of  material 
of  an  external  force  applied  to  it,  and  we  will  choose  for  our 
investigation  a  straight  rod  of  uniform  cross-section  and  will 
not  consider  the  force  of  gravity  as  acting.  We  will  apply 
to  the  end  of  our  rod  a  pull,  F  (not  sufficient  to  break  or  per- 
manently change  its  shape),  which,  in  order  that  the  rod 
remain  stationary,  will  require  an  equal  pull  in  the  opposite 
direction  at  the  other  end.  These  forces  tend  to  tear  apart 
the  particles  of  the  material,  and  as  the  bar  remains  intact 
the  particles  must  be  in  a  different  condition,  relative  to  each 
other,  from  that  in  which  they  were  before  we  applied  the 
pull.  If  instead  of  a  pull  we  exert  a  pressure  on  one  end,  an 
equal  and  opposite  pressure  must  be  applied  to  the  other  end 
to  keep  the  bar  in  equilibrium,  and  the  particles  of  the 
material  will  now  tend  to  crowd  together  and  crush  each 
other.  In  both  of  these  cases  we  have  arranged  our  forces  so 
that  the  bar  does  not  move  and  they  have  been  taken  small 
enough  so  as  not  permanently  to  change  its  dimensions.  Now 
as  the  length  of  the  bar  separates  the  points  of  application  of 
our  forces,  there  must  be  some  action  set  up  among  the  par- 
ticles of  the  bar  itself  which  transmits  the  force  from  one  end 


Pull 


H H 


In  tension 
Pressure 


In  compression 
Fig.  1.  Fig.  2. 

to  the  other,  or  to  a  common  point  of  action.  Let  us  now 
imagine  a  plane  passed  through  the  bar  perpendicular  to  its 
axis:  in  the  first  case  (the  pull,  Fig.  1)  our  forces  would  tend 


TENSION  AND  COMPRESSION.  5 

to  pull  apart  the  two  pieces  of  the  bar;  in  the  second  (not 
shown)  they  would  press  them  together,  each  piece  would 
tend  to  move  in  the  direction  of  the  external  force  acting  on 
it  and  the  amount  of  this  tendency  would  be  equal  to  that 
force  so  that  the  total  action  in  the  bar  between  the  particles 
on  either  side  of  any  imaginary  section  would  be  equal  but 
opposite  to  the  external  forces  applied.  The  action  of  all  the 
particles  on  the  right  side  of  any  section  would  be  equal  to 
but  opposite  to  the  force  on  the  right  end,  and  of  those  on  the 
left  side  equal  to  but  opposite  to  the  force  on  the  left  end. 
This  must  be  true  in  order  that  the  bar  remain  intact,  i.  e.  in 
equilibrium. 

3.  Stress.  —  When  any  such  action  as  the  above  is  set  up 
among  the  particles  of  a  piece  of  material  that  piece  is  said 
to  be  under  stress,  and  the  external  force  which  causes  the 
stress  is  called  the  load.     A  rod  under  the  action  of  a  pull  is 
said  to  be  in  tension,  and  the  pull  is  the  tensile  load.     A  rod 
to  which  pressure  is  applied  is  said  to  be  under  compression, 
and  the  pressure  is  called  the  compressive  load.     Hereafter 
in  using  the  word  stress  we  will  mean  the  amount  of  action 
between  the  particles  in  unit  cross-sectional  area,  and  will 
use  " total  stress"  for  the  action  over  the  area  of  the  whole 
cross-section.     The  stress  per  unit   cross-sectional   area  is 
sometimes  called  " intensity  of  stress"  and  is  equal  to  the 
external  force,  F,  divided  by  the  cross-sectional  area,  A.     The 
forces  on  one  side  of  the  section  only  are  used,  so, 

F 

p  (stress)  =  —  • 
A. 

Of  course  we  would  get  the  same  result  by  using  the  forces 
on  the  other  side  of  the  section,  as  the  bar  must  be  in  equi- 
librium; they,  however,  would  act  in  the  opposite  direction. 

4.  Now  let  us  see  what  will  happen  if  we  gradually  increase 
the  load,  F.     We  know  that  all  materials  are  more  or  less 
elastic,  so  as  F  is  increased  the  bar  will  stretch  or  shorten 


6  STRENGTH   OF  MATERIAL. 

according  as  F  puts  it  in  tension  or  compression;  and  up  to  a 
certain  value  of  F,  the  bar  will  spring  back  to  its  original 
length  when  F  is  removed.  If  we  experiment  carefully  with 
any  material  we  will  find  that  there  is  for  it  a  certain  value  of 
F,  after  reaching  which,  and  then  having  removed  F,  the  bar 
will  be  found  a  little  longer  or  a  little  shorter  than  it  was 
originally;  in  other  words,  it  will  have  a  "permanent  set." 
This  value  of  F  for  unit  sectional  area  is  called  the  "  limiting 
stress"  or  "elastic  limit"  of  that  material.  It  is  obvious 
that  no  part  of  our  machine  or  structure  must  be  subjected  to 
a  force  equal  to  this,  for  if  it  be  the  piece  so  used  is  afterward 
unfit  to  do  its  work. 

If  we  go  on  increasing  F  the  bar  will  finally  break  or  crush, 
but  at  present  we  are  interested  only  in  the  elastic  limit,  the 
stress  for  which  we  will  call  /,  and  for  several  materials  its 
value  will  be  found  in  the  table  at  the  end  of  this  chapter. 

F 
We  know  that  any  stress,  p,  is  equal  to  — ,  so  to  find  the 

A. 

tensile  or  compressive  load  any  piece  of  material  can  sustain 

F 

unhurt,  we  put/  =  —  or  F  (the  limiting  load)  =  fA.     This 
A 

is  true  for  all  ordinary  lengths  of  material  under  tension,  and 
for  short  pieces  under  compression.  We  will  later  (Chapter 
XI)  consider  long  pieces  under  compression,  in  which  the 
stress  due  to  bending 'must  be  taken  into  account. 

5.  In  the  preceding  article  we  saw  that,  within  the  elastic 
limit,  a  piece  of  material  would,  when  the  load  was  removed, 
return  to  its  original  length.  After  voluminous  experiments, 
it  has  been  found  that  within  the  elastic  limit  the  extension 
or  compression  of  a  rod  under  stress  varies  directly  as  the 
stress,  or  stress  is  equal  to  a  constant  multiplied  by  the  ex- 
tension or  compression.  This  is  known  as  Hooke's  Law,  from 
its  discoverer.  If  we  let  x  denote  the  total  extension  of  a 

rod  of  length  I,  the  extension  per  unit  length  will  be  — ,  and 


TENSION  AND  COMPRESSION.  7 

this  extension  per  unit  length  is  called  strain,  and  we  will 
denote  it  by  e;  then  if  p  denote  the  stress  per  unit  area  of 
cross-section, 


E  being  a  constant  found  by  experiment  and  called  the 
modulus  of  elasticity.     Its  value  for  several  materials  will  be 

found  in  the  table  at  the  end  of  this  chapter,  E  =  —  • 

e 

6.    Work  Done  on  a  Piece  of  Material  by  a  Load.  —  In  all 

cases  of  overcoming  resistance  work  is  done,  and  we  will  now 

find  how  much  work  is  done  on  a  piece  of  material  on  which 

a  load  is  acting.     By  Hooke's  Law  p  =  Ee  so  that,  as  we  have 

equilibrium  within  the  elastic  limit,  the  resistance  of  the 

material  to  the  load  must  equal  p  per  unit  area.     Before  the 

load  is  applied  the  resistance  is  nothing.     Now  let  us  apply 

the  load  very  slowly.      The  re- 

sistance  will   increase   with    the 

load   from  zero  to  the  equal   of 

the  final  load.    It  follows  that  the 

mean  resistance  will  be  equal  to 

one-half  of  the  final  load.     Work 

is     equal    to     force     multiplied  Fig.  3. 

by    the    space    through    which 

it  acts.     Due  to  the  load,  our  rod  has  stretched  through  the 

distance  x.     Hence  the  work  done  is  equal  to  one-half  the 

final  resistance  multiplied  by  x,  or  if  L  be  the  final  load 

L 

the  work  is  equal  to  —  X  x. 
& 

*-¥• 

The  following  graphic  method  may  be  clearer:    Let  the 
abscissa  represent  the  extensions,  and  the  ordinates  the  loads 


8  STRENGTH  OF   MATERIAL. 

causing  them.  When  x  =  0,  L  =  0,  and  as  the  load  in- 
creases so  will  the  extension  in  accord  with  the  formula 
p  ==  Ee,  giving  us  a  straight  line  for  the  load  curve.  The 
work  done  will  now  be  represented  by  the  area  between  the 
load  curve,  OA,  the  axis  of  x,  and  the  ordinate,  Ab,  represent- 
ing the  final  load,  L. 

extension 
Work  = X  L  =  area  of  triangle  OAb. 

The  following  is  the  method  by  calculus: 

L  (load)  =  pA  (Art.  4), 

p  =  E~  =  Ee  (Art.  5). 

The  load  curve  then  is 

L  =  pA  =  AE  j=AEe. 

The  work  done  by  any  load  L  acting  through  a  space  dx  is 

Ldx  or  dW  =  Ldx, 
but 


which  integrated  between  the  limits  for  x  gives 

EA          X2  ~]  total  extension 

'      U 

which  as  p  =•-  —  and  L  =  pA  gives  W        —  for  the  work 
as  before. 


TENSION  AND  COMPRESSION.  9 

If  our  bar  be  stretched  to  just  within  the  elastic  limit  we 
will  have 


also 


/=        orL  =fA; 


hence,  from  the  above, 

f2Al       f2 

W  =  —-  =  •£-    XF.       (V  equals  volume  of  bar). 
2h       2h/ 

As  we  have  stretched  our  bar  to  just  within  the  elastic 
limit,  it  will  return  to  its  original  length  when  we  remove  the 
load.  This  load  then  does  the  greatest  amount  of  work 
that  can  be  done  on  a  piece  of  material  without  injuring  it. 
While  stretched,  the  rod  has  stored  in  it  an  amount  of 
energy  equal  to  the  work  done  in  stretching  it  to  its  elastic 
limit.  This  stored  energy  is  called  the  resilience  of  the  bar, 

/2 

and  the  part  —  —  is  the  modulus  of  resilience. 
2E 

7.  The  work  done  by  forces  quickly  applied  is  much 
greater  than  if  the  force  is  slowly  increased,  for,  suppose  a 
vertical  rod  having  a  collar  round  its  lower  end  is  stretched 
by  a  weight  W  falling  from  a  height  h  upon  the  collar. 
The  work  done  by  the  falling  weight  is  W  (h  +  x)  and  this 

Lx 
must  equal  the  work  done  on  the  rod  or  —  •    We  have  then 


or  the  slowly  applied  load  which  would  stretch  the  rod  to 
the  same  extent  as  the  falling  weight  would  have  to  be  con- 
siderably greater  than  the  weight. 


10  STRENGTH  OF  MATERIAL. 

Again,  if  we  imagine  a  load  L  to  be  instantaneously 
applied  and  to  cause  a  strain  equal  to  x  and  a  load  L, 
slowly  applied  which  would  cause  the  same  strain,  the  work 
done  in  the  first  case  or  Lx  (force  times  space)  must  equal 

T    ~ 

the  work  done  in  the  second   case  or  -  —  •      Equating  we 

have  Lt  =  2L  or  a  suddenly  applied  load  has  twice  the  effect 
of  a  slowly  applied  one. 


Definitions: 

STRESS.  —  When  due  to  the  load  on  a  body,  there  is 
mutual  action  between  the  particles  on  either  side  of  a  sec- 
tion through  the  body,  so  that  the  particles  on  one  side  exert 
a  force  on  those  on  the  other  side,  stress  is  said  to  exist  in 
that  body,  and  the  intensity  of  the  stress  is  the  force  per  unit 
area  of  cross-section.  Briefly,  then,  stress  is  force  per  unit 
area  of  cross-section. 

STRAIN.  —  The  ratio  of  change  of  length  in  a  body  due  to 
the  load  on  it,  to  the  original  length;  briefly,  change  of 
length  per  unit  length. 

MODULUS  OF  ELASTICITY.  —  The  stress  which  would 
double  the  length  of  a  rod  provided  Hooke's  Law  held  good 

/load\ 

for  that  extension:   or,  it  is  the  ratio  of  unit  stress  I  -   -    to 

\area/ 

/total  extension\ 

unit  strain  I  -  ;    or  ratio  of   stress  to  strain 

\original  length  / 

within  Hooke's  Law. 

ELASTIC  LIMIT.  —  That  stress,  which  if  exceeded  will  pro- 
duce a  permanent  change  of  length;  or  the  maximum  stress  a 
material  can  suffer  without  being  permanently  deformed. 

RESILIENCE.  —  When  a  bar  is  loaded  to  its  elastic  limit, 
the  work  done  in  stretching  it  is  called  the  resilience  of  the 
bar,  and  the  modulus  of  resilience  is  this  work  divided  by  the 


TENSION  AND   COMPRESSION. 


11 


volume  of  the  bar;  or,  resilience  is  the  capacity  of  a  body 
to  resist  external  work. 

ULTIMATE  STRENGTH  is  the  stress  which  produces  rupture. 

WORKING  LOAD  is  the  maximum  stress  to  which  a  piece 
of  material  will  be  subjected  in  actual  practice. 


STRENGTH    OF   MATERIAL 


Average  values  in  pounds  per  square  inch. 

Modulus 

Material. 

Weight 
per 
cubic 
foot. 

Elastic 
limit. 

/ 

Modulus 
of 
elasticity. 

of 
elasticity 
for 
shear  and 

Ultimate 
fiber 

stress. 

Limiting 
shearing 
stress. 

Pounds. 

E. 

torsion. 

q 

C. 

Steel  

490 

I  35,000  to  ) 
|  50,000      j 

29,000,000 

10,500,000 

110,000 

50,000 

Iron,  cast   .... 

450 

i    6,000  T) 
\  20,000  C  } 

15,000,000 

5,000,000 

35,000 

7,850 

Iron,  wrought 

480 

25,000 

25,000,000 

10,000,000 

54,000 

20,160 

Brass,  cast     .  . 

9,500,000 

20,000 

4,030 

Brass,  drawn 

520 



14,500,000 

5,600,000 

70,000 

Copper,  cast  . 
Copper,  drawn 

540 

12,000,000 
15,000,000 

'6,000,000 

22,000 
65,000 

2,890 

Stone,  granite. 

'l60 

2,000  ' 

6,000,000 

1,800,000 

2,000 

Timber  

40 

3,000 

1,500,000 

140,000 

10,000 

'   1,200 

Examples: 

1.  A  steel  bar,  5  ins.  long,  sectional  area  \  sq.  in.,  stretches 
.007  in.  under  a  load  of  20,000  Ibs.,  and  shows  no  permanent 
elongation  when  the  load  is  removed.     What  is  the  modulus 
of  elasticity  of  the  metal? 

Solution: 

Ft       20,000  X  5  X  2 

E  =  -  — —         =  28,571,428.571  in.-lbs. 

A.X  .1)07 

2.  A  vertical  wrought-iron  rod   200  ft.   long  has  to  lift 
suddenly  a  weight  of  2  tons.      What  is  the  area  of  its  cross- 
section  if  the  greatest  strain  to  which  wrought  iron  may  be 


12  STRENGTH  OF  MATERIAL. 

subjected  is  .0005  for  unit  length.    E  =  30,000,000.    Neglect 
the  weight  of  the  rod.     (See  Art.  7.) 

F 

Solution:    Stress  =  8960  Ibs.       —  =  p  =  Ee. 

A. 

F  8960 

=  E~e  =  30,000,000  X. 0005  °r  A  =  ^  Sq'  in" 

3.  Find  the  area  of  cross-section  in  example  2,  taking 
the  weight  of  the  rod  into  account. 

Ans.     .611  sq.  in.,  or  .625  if  rod  also  is  suddenly  raised. 

4.  A  steel  rod  J  sq.  in.  in  sectional  area  and  5  ft.  long  is 
found  to  have  stretched  Tfo  in-  under  a  load  of  1  ton.     What 
is  the  modulus  of  elasticity  of  steel? 

Ans.    35,840,000  in.-lbs. 

5.  A  chain  30  ft.  long  and  sectional  area  }  sq.  in.  sus- 
tains a  load  of  3900  Ibs.;  an  additional  load  of  900  Ibs.  is 
suddenly  applied.     Find  the  resilience  at  the  instant  the 
900  Ibs.  is  applied.     E  =  25,000,000. 

Ans.     25.992  ft.-lbs. 

6.  A  steel  rod  3|  ft.   long,   2£  sq.   ins.   sectional  area, 
reaches  the  elastic  limit  at  125,000  Ibs.,  with  an  elongation 
of  .065  in.     Find  the  stress  and  strain  at  the  elastic  limit, 
the  modulus  of  elasticity,  and  the  modulus  of  resilience  of 
steel,  and  express  each  in  its  proper  units. 

Ans.  /  =  50,000  Ibs.  per  sq.  in.;  e  =  .00166§;  ^E  = 
30,000,000  Ibs.  per  sq.  in.;  and  modulus  resilience  =  41§ 
Ibs.  per  sq.  in. 

7.  A  piston  rod  is  10  ft.  long  and  7  ins.  in  diameter. 
The  diameter  of  the  cylinder  is  5  ft.  10  ins.,  and  the  effective 
steam  pressure  is  100  Ibs.  per  sq.  in.     Find  the  stress  pro- 
duced and  the  total  alteration  in  length  of  the  rod  for  a 
complete  revolution.     E  =  30,000,000. 

Ans.     Change  in  length  .0796  in. 


TENSION  AND  COMPRESSION.  13 

( 

8.  How  much  work  can  be  done  in  stretching  a  com- 
position rod  5  ft.  long  and  2  in.  in  diameter,  without  injury, 
if  the  proof  stress  of  the   metal  is  2.8  tons  per  sq.   in.? 
E  =  4928  ton-ins. 

22 

Ans.     .15  ton-ins.,  using  x  =  —  • 

9.  The    proof   strain    of   iron    being   TTnrc,   what  is   the 
shortest  length  of  rod  1^  sq.  ins.  sectional  area,  which  will 
not  take  a  permanent  set  if  subjected  to  the  shock  caused 
by  checking  the  weight  of  36  Ibs.  dropped  through  10  ft., 
before  beginning  to  strain  the  rod?     E  =  30,000,000. 

Ans.     192.31  ins. 

10.  Find  the  shortest  length  of  steel  rod,  2  sq.  ins.  sec- 
tional area,  which  will  just  bear,  without  injury,  the  shock 
caused  by  checking  a  weight  of  60  Ibs.  which  falls  through 
12  ft.  before  beginning  to  strain  the  rod.     E  =  30,000,000. 
Modulus  of  resilience  =  15  Ib.-ft. 

Ans.     24.05  ft. 

11.  A  brass  pump  rod  is  5  ft.  long  and  4  ins.  in  diameter 
and  lifts  a  bucket  28  ins.  in  diameter,  on  which  is  a  pressure 
of  6  Ibs.  per  sq.  in.,  in  addition  to  the  atmosphere,  against  a 
vacuum  below  the  bucket  which  reduces  the  atmospheric 
pressure  to  2  Ibs.     What  is  the  stress  in  the  rod  and  the 
total  extension  per  stroke?     E  =  9,000,000. 

Ans.     Stress  925  Ibs.  per  sq.  in.,  and  extension  .0061  in. 

12.  Assuming  a  chain  twice  as  strong  as  the  round  bar  of 
which  the  links  are  made,  what  size  chain  must  be  used  on 
a  20-ton  crane  with  three  sheaved  blocks  if  /  =  6000? 

Ans.     Diameter  of  section  of  metal  .8899  in. 

13.  A  piston  rod  is  9  ft.   long  and  8  ins.  in  diameter. 
The  diameter  of  the  cylinder  is  88  ins.  and  the  effective 
pressure  is  40  Ibs.  per  sq.  in.     What  is  the  stress  produced 


14  STRENGTH  OF  MATERIAL. 

and  the  total  alteration  in  length  of  the  rod  per  revolution? 
E  =  29,000,000. 
Ans.     x  =  .0359  in. 

14.  Find  the  work  done  in  Example  13,  and  find  the  resil- 
ience of  the  rod  if/  =  12  tons. 

Ans.     W  =  3.88  in.-tons.     R  =  52.5  in.-tons. 

15.  The  stays  of  a  boiler  in  which  the  pressure  is  245  Ibs. 
per  sq.  in.  are  spaced  16  ins.  apart.     What  must  be  their 
diameter  if  the  stress  allowed  is  18,000  Ibs.  per  sq.  in.? 

Ans.     2J  ins. 

16.  What  is  the  area  of  the  section  of  a  stone  pillar  carry- 
ing 5  tons  if  the  stress  allowed  is  150  Ibs.  per  sq.  in.? 

Ans.     75  sq.  ins. 

17.  What  is  the  length  of  an  iron  rod  (vertical)  which  will 
just  carry  its  own  weight?    /  =  9000  Ibs.  per  sq.  in. 

Ans.     2700  ft. 

18.  The  coefficient  of  expansion  of  iron  is  .0000068  in.  per 
degree  F.     An  iron  bar  18  ft.  long,  1£  ins.  in  diameter,  is 
secured  at  400°  F.,  between  two  walls.     What  is  the  pull  on 
the  walls  when  the  bar  has  cooled  to  300°  F.? 

Ans.     34,862  Ibs. 

19.  Coefficient   of   expansion   of   copper   is   .0000095   in. 
E  =  17,000,000.     A  bar  of  iron  is  secured  between  two  bars 
of  copper  of  the  same  length  and  section  at  60°  F.     What  are 
the  stresses  in  the  bars  at  200°  F.? 

Ans.     In  copper  2958,  iron  5916  Ibs.  per  sq.  in. 

20.  A  bar  of  iron  is  3  ft.  long,  2  ins.  in  diameter.     The 
middle  foot  is  turned  down  to  one  inch  diameter.     Compare 
the  resilience  with  the  original  bar  and  with  a  uniform  bar 
of  the  same  weight. 

Ans.     (1)  1  to  8;  (2)  1  to  6. 


CHAPTER  II. 


SHEARING. 

8.  We  will  next  apply  force  to  produce  shearing.    In  study- 
ing shearing  we  will  use  the  same  sort  of  rod  we  used  in  the 
preceding  chapter  but  will  apply  our  force  in  a  different  way. 
To  get  the  proper  effect  we  will  slip  our  rod  through  holes 
bored  in  two  extra  pieces  of  material  as  in  Fig.  4,  and  then 
apply  opposite  and  equal   pulls  to   the 

two  extra  pieces  so  that  the  effect  on  the 
rod  will  be  to  pull  one  part  up  and  the 
other  down.  If  we  pull  hard  enough  our 
rod  will  be  sliced  off  smoothly  as  though 
a  plane  had  been  passed  through  it  per- 
pendicular to  its  axis.  It  will  be  sheared 
off.  Practically  this  is  as  near  as  we  can 
approximate  to  pure  shearing.  In  theory 
the  two  forces,  F,  should  act  on  either 
side  of  a  section  of  the  rod  in  parallel  planes  indefinitely 
close  to  each  other  so  that  their  pull  would  induce  no 
tendency  to  turning  or  bending;  then  whether  we  actually 
sheared  the  bar  through  or  not,  the  stress  in  the  bar  on  one 
side  of  the  section  would  be  the  force  F  divided  by  the 

F 

area  A,  of  the  section  of  the  bar,  or,  q  =  —  - 

A. 

It  will  be  noticed  that  the  direction  of  the  shearing  stress 
is  parallel  to  the  section,  while  in  tensile  or  compressive  stress 
the  direction  is  normal  to  the  section ;  for  this  reason  shearing 
is  sometimes  called  tangential  stress. 

9.  Two  plates  bearing  a  longitudinal  load  and  held  together 
by  a  riveted  joint  present  a  good  illustration  of  shearing. 

15 


16 


STRENGTH  OF  MATERIAL. 


The  rivet  is  under  almost  pure  shear  if  it  closely  fits  the  rivet 
holes,  and  the  bearing  surfaces  of  the  plates  are  plane.     Fig.  5 

shows    a   section    and    plan    of 

I    [I  a  single-riveted  lap  joint.     The 

• j_>  *  distance  between  the  centers  of 

the  rivet  holes  is  called  the 
pitch  p,  and  it  is  obvious  that 
each  rivet  supports  the  load  on 
a  strip  of  plate  equal  in  width 
to  the  pitch  of  the  rivets.  So 
that  if  F  is  the  force  acting 
Fig.  5.  along  this  strip  of  plate,  the 

F 

shearing  stress,  q,  on  the  rivet  supporting  it  is  equal  to  — 

A. 

as  before,  or  as  A  is  —  ,  where  d  is  the  diameter  of  the  rivet, 


—  ^ 

-o 

—  

F*- 

?©; 

—  F 

1  •^ 

O 

we  have  for  the  shearing   stress   q  =  — .     If  we   put  the 


limiting  shearing  stress  for  the  material  of  which  our  rivet  is 
made  for  q,  we  can  find  the  diameter  of  the  rivet  we  must  use 

under  the  conditions,  d  =  2  V  —  • 

nq 

In  working  joints,  such  as  pin  joints,  the  action  is  not  pure 
shear,  but,  due  to  the  clearance  necessary  for  a  working  fit, 
there  is  some  bending.      Ex- 
periment   has    proved    that 
the  stress   at  the    center  of 
the  pin  of  such  a  joint  is  $ 
of  that  found  by  the  above 
formula.     To   find   the   dia-  Fig  6 

meter  of  the  pin    necessary 

for  a  joint  like  that  in  Fig.  6,  we  first  notice  that  there  are 
two  sections  of  the  pin  to  be  sheared, 


SHEARING. 


17 


then, 


,  _  I  shearing  stress 
^       I  at  center  of  pin 


4        F 
3  "  2~A 


SF 


or, 


d  = 


The  limiting  shearing  stress  for  several  materials  is  given  in 
the  table  at  the  end  of  Chapter  I. 

10.  Usually  the  value  of  F  for  the  above  is  readily  found, 
as  the  thrust  on  a  piston  rod,  etc.;  but  for  the  force  acting  on 
riveted  joints  the  ordinary  steam  boiler  will  give  us  a  good 
example.  Fig.  7  is  a  section  through  a  boiler  designed  to 
carry  a  steam  pressure  of  s 
pounds  per  square  inch,  its 
radius  is  r,  and  we  wish  to 
find  the  tangential  force  at 
any  point  A .  Draw  the  per- 
pendicular diameters  AB 
and  CD,  then  on  a  ring  1 
inch  wide  there  will  be  a 
pressure  of  s  pounds  on  each 
inch  of  the  circumference. 
Now  if  we  should  resolve 


Fig.  7. 


horizontally  and  vertically 
the  pressure  on  each  one  of 
these  square  inches,  the  sum  of  all  the  horizontal  components 
would  be  zero.  If,  however,  we  take  the  sum  of  all  the  horizon- 
tal components  on  the  right  half  of  the  ring  we  will  get  all  the 
forces  which  act  toward  the  right;  of  course  those  to  the  left 
of  the  diameter  A B  will  be  equal  but  opposite  in  direction. 
It  will  be  further  noticed  that  to  support  these  forces  we  have 
the  boiler  material  at  A  and  also  at  B  and  both  these  points 
support  equal  "amounts,  so  we  can  divide  the  force  acting  on 
the  semicircle  by  2  to  get  that  part  which  acts  at  A  or  5;  or, 


18  STRENGTH  OF   MATERIAL. 

we  can  take  the  sum  of  the  horizontal  components  of  the 
pressure  on  that  part  of  the  ring  from  C  to  A  which  will  give 
the  same  result.  Finding  this  sum  is  most  readily  done  by 
calculus.  Take  any  point  on  the  circumference  and  let  its 
angular  distance  from  C  be  0,  then  rdd  (remembering  our 
strip  is  1  inch  wide)  will  be  an  element  of  area  on  which  the 
pressure  is  s  X  rdd  acting  radially  and  of  which  the  horizontal 
component  is  srdO  cos  6,  and  if  we  integrate  this  expression 

between  the  limits  0  and  -  we  will  get  the  sum*  of  all  these 
components,  or  the  force  H. 

H  =  sr   I  z   cos  Odd  =  sr   sin  0  p  = 


=  sr. 


Having  H,  the  force  on  a  strip  1  inch  wide,  the  force  on  a 
strip  supported  by  one  of  pur  rivets  is  found  by  multiplying 
this  by  the  pitch,  or  F  =  psr,  so  that  the  shearing  stress  on 
the  rivet  is 

psr 


The  vertical  components  have  no  tensile  effect  at  A,  but  form 
all  the  tensile  effect  at  C. 

11.  In  this  connection  we  can  find  the  required  thickness 
of  a  cylindrical  shell,  such  as  a  steam  pipe,  remembering  that 
it  is  under  tensile  stress,  not  shearing,  and  changing  our  con- 
stants accordingly.  If  I  is  the  length  considered  and  t  the 
thickness  of  the  plate,  the  sectional  area  of  the  metal  will 
be  It,  and  if  /  is  the  tensile  strength  allowed,  the  resistance 
of  the  shell  will  befit;  this  must  be  balanced  by  the  force  IF, 
which  from  Art.  10  is  Isr: 

sr 
so  fit  =  Isr;  or  t  =  —  • 


SHEARING.  19 

If  we  wish  the  thickness  of  boiler  plates,  we  must,  since  boilers 
are  built  of  plates  with  riveted  joints,  divide  our  result  by 
the  efficiency  of  the  joint,  strength  of  joint  being  equal  to  the 
strength  of  the  solid  metal  multiplied  by  the  efficiency  of  the 
joint. 

12.  We  have  seen  how  to  find  the  shearing  stress  on  a  rivet 
of  a  single-riveted  lap  joint,  but  we  must  remember  we  have 
taken  out  of  our  plate  a  piece  of  metal  equal  to  the  diameter 
of  the  rivet  hole,  so  that  we  have  left  in  the  strip  of  plate,  to 
bear  the  whole  force  F,  a  section  whose  area  is  (p  —  d)  t,  and 
the  tensile  stress  on  this  section  must  not  exceed  the  tensile 

F 

strength    of    the    material,    or   to    balance;  /  =  —       —  —  ; 

t(p  —  d) 

transposing,  the  pitch  for  a  single-riveted  la~p  joint  is 


There  is  another  kind  of  joint  called  the  butt  joint,  where  a 
narrow  strip  of  material,  or  strap,  covers  the  edges  of  the 


1 

J 

••«— 

1 

Fig.  8. 


\ 

i 

i  — 

I 

u 

i 

plates  and  is  riveted  to  both.  This  joint  is  called  a  single 
butt  joint,  single-riveted  (Fig.  8),  and  is  treated  exactly  as 
a  single-riveted  lap  joint. 

If  there  are  two  straps  used,  one  on  either  side  of  the  plates 
as  in  Fig.  9,  the  joint  is  called  a  double  butt  strap  joint, 


20 


STRENGTH  OF   MATERIAL. 


single-riveted,  and  here  it  will  be  noticed  we  have  two  sec- 

F 

tions  of  rivet  to  shear,  so  our  formula  becomes  q  = 

2  A 

Now  suppose  we  have  two  rows  of  rivets  in  either  a  lap  joint 
or  single  butt  joint  (where  more  than  one  row  is  used  the 
rivets  are  generally  staggered  as  in  Fig.  10,  shown  in  the 


o 

o 

0 

0 

o 

o 

0 

o 

0 

0 

0 

0 

o 

o 

0 

0 

0 

1 

o 

1  — 

Fig.  10. 

plan)  we  will  have  two  sections  of  rivets  to  shear,  and  our 
formula  will  be  the  same  as  for  the  single-riveted  double  butt 
strap, 

F 


and  if  we  have  for  these  joints  n  rows  of  rivets,  the  formula 
becomes 

F 


If  we  use  double  butt  straps  we  will  have  double  the  number 
of  sections  to  shear,  so  have  to  divide  the  above  value  of  F  by 
2;  or,  for  double  butt  straps, 

F 


SHEARING. 


21 


Single-riveted  . 
Double-riveted  . 

Lap  joint  or 
single  butt  strap. 
F                   qxd* 

Double  L 

q  —       ,.,    or  I<  —  -  —  -.  — 
Tra                         4 

4 

7T  Cl 

2^T 

F      _  v      nqxd2 

^ 

-4>rf 

Q 

or  7'1 


or 


2  nqxd* 


13.  If  we  have  more  than  one  row  of  rivets  the  plate  will, 
if  it  be  ruptured,  carry  away  along  the  outer  row  of  rivets 
in  every  case;  for  (see  Fig.  11)  if  it  did  not,  there  would  be 


Fig.  11. 

one  or  more  rows  of  rivets  to  be  sheared  in  addition  to  the 
carrying  away  of  the  plate,  and  clearly  the  rupture  would 
occur  where  it  would  require  the  least  force.  Consequently, 
as  far  as  the  stress  on  the  plate  is  concerned,  it  may  always 
be  computed  as  we  have  it  in  the  final  part  of  Art.  12,  using 
the  limiting  stress  allowed  for  the  material;  or, 

f(p-d)t  =F. 

At  the  end  of  Art.  12  we  have  formula  giving  F  for  the  shear- 
ing stress  of  the  material  of  the  rivets,  and  if  we  equate  these 
two  values  of  F  we  can  find  the  pitch  of  our  rivets  for  equal 
strength  (that  is,  rupture  will  be  as  likely  to  occur  by  shearing 
the  rivets  as  by  the  plate  carrying  away),  if  we  know  their 
diameter  and  the  thickness  of  the  plates. 


22  STRENGTH  OF   MATERIAL. 

Equating  we  have 


which  gives  the  pitch  where  particular  plates  and  rivets  are 
to  be  used. 

In  these  formulae 

p  =  pitch  of  rivets, 

d  —  diameter  of  rivets, 

n  =  number  of  rows, 

q  =  limiting  shearing  stress  for  rivets, 

/  =  limiting  tensile  stress  for  plate, 

t  =  thickness  of  plate. 

14.  When  a  bar  is  subjected  either  to  tension  or  compres- 
sion there  is  shearing  stress  along  any  oblique  section.  Let 
the  angle  BDC  =  0,  then  the  intensity  of  the  force  F  (acting 

F 

in  either  direction)  along  the  section  DB  is  -       —  -  ,  but  the 

area  DB 

area  DB  equals  the  area  of  BC  X  cosec  0,  or  if  A  is  the 


D     c 

Fig.  12 

area  of  a  right  section,  the  intensity  of  the  force  on  DB  is 

F 

— .  The  component  of  this  force  resolved  along  DB  is 
A  cosec  0 

F  F  F 

-cos  d  =  —  sin  0  cos  0  =-  -  sin  (2 0).     This  being 
A  cosec  0  A  2  A 

F 

a  tangential  force  is  shearing  stress.  —  resolved  per- 

A  cosec  0 


SHEARING.  23 

F  F 

pendicular  to  DB  is-        sin  0  =— -sin2#,  the  normal 

A  cosec  0  A 

stress.     When  0  =  0  or  -  the  shearing  stress  is  0,  but  for  any 

£i 

other  value  of  6  it  is  a  finite  quantity,  which  proves  our  propo- 
sition.    When  0  =  0  the  normal  stress  is  0,  but  increases  with 

d  to  a  maximum  when  0  =  - ,  which  is  as  it  should  be. 

2 


Examples : 

1.  Two  wrought-iron  plates,  3  ins.  wide  by  J  in.  thick  are 
lap-jointed  by  a  single  rivet,  1  in.  in  diameter.     What  will  be 
the  pull  required  to  break  the  joint,  if  the  tensile  strength 
is    18    tons  per  sq.    in.?      What  is   the   efficiency  of   the 
joint? 

Solution:  Area  (plate)  =3Xi--lXi  =  lsq.  in.4 
.'.  Strength  =  18  tons.  Area  of  section  without  rivet  hole 
=  |  sq.  in.  /.  Strength  =  |  X  18  =  27  tons.  Efficiency 

of  the  joint  is  ||  or  §  =  66§%.     Area  of  rivet  section  =  -• 

4 

If  it  is  of  the  same  material  as  the  plate  the  force  necessary 

and2       18  X  22 

to  shear  it  is  F  = =  -          -  —  14.13  -f  tons  and  the 

4  4X7 

joint  would  break  by  shearing  the  rivet. 

2.  A  cylindrical  vessel  with  hemispherical  ends,  diameter 
6  ft.,  is  exposed  to  internal  pressure  200  Ibs.  above  the  atmos- 
phere.    It  is  constructed  of  solid  steel  rings  riveted  together. 
If  /  =  7  tons  per  sq.  in., 'how  thick  must  the  metal  be,  and 
what  is  the  longitudinal  stress  in  the  metal  of  the  ring  joint 
whose  section  is  ^  that  of  the  solid  plate? 

sr       200  X  6  X  12 
Solution:  e==      =  =  .459  in. 


24  STRENGTH  OF  MATERIAL. 

.,     ,.     ,  10  nr2p      5rp 

Longitudinal  stress  =  7rr2p  =  ^  .  2  xrtq,  or,  q  =  — 

7.2  irrt         1 1 

5  X  3  X  12  X  200 
q=   7  X. 459X2240 

3.  What  must  be  the  thickness  of  a  copper  pipe,  j  in.  in 
diameter,   to  sustain  a  pressure  of   1350  Ibs.  per  sq.  in.? 
/  =  950  Ibs.  per  sq.  in. 

Ans.     .533  in. 

4.  A  single-riveted  lap  joint,  plate  J  in.  thick,  is  under  a 
load  of  3  tons  per  sq.  in.  of  plate  section.     Rivets  f  in.  in 
diameter,  pitch  1J  ins.     What  is  the  shearing  stress  on  the 
rivets  and  the  efficiency  of  the  joint? 

Ans.     Shearing  stress  3.82  tons.     Efficiency  60%. 

5.  Two  plates,  f  in.  thick,  are  double-riveted  with  double 
butt  straps,  rivets  1  in.  in  diameter.     The  shearing  strength 
of  the  rivets  is  f  the  tensile  strength  of  the  plates.     Find  the 
pitch  and  the  efficiency  of  the  joint. 

Ans.     Pitch  4£  ins.     Efficiency  .7778. 

6.  What  is  the  pitch  of  the  rivets  in  a  treble-riveted,  double 
butt  strap  joint  between  plates  J  in.  thick,  rivets  f  in.  in 
diameter,  if  the  tensile  stress  of  the  plates  is  limited  to  10,000 
Ibs.   per  sq.   in.,  and  the  shearing  stress  of  the  rivets  to 
8000  Ibs.  per  sq.  in.?     What  is  the  efficiency  of  the  joint? 

Ans.     Pitch  f  f  $  in.     Efficiency  83  +  %. 

7.  A  cylindrical  boiler  8  ft.  4  ins.  in  diameter  is  under  100 
Ibs.  per  sq.  in.  pressure.     What  must  be  the  thickness  of  the 
plates  that  the  stress  may  not  exceed  4000  Ibs.  per  sq.  in.? 

Ans.     1J  ins. 

8.  What  is  the  pitch  of  1-in.  rivets  in  a  double-riveted  lap 
joint  between  |-in.  steel  plates?    Tensile  stress  13.2  tons  and 
shearing  stress  10.5  tons  per  sq.  in.     What  is  the  efficiency  of 
the  joint? 

Ans.     Pitch  3£  ins.     Efficiency  71+  %. 


SHEARING.  25 

9.  The  steel  plates  of  a  girder  are  J  in.  thick,  treble-riveted 
with  double  butt  strap,  with  1-in.  rivets.     Shearing  stress  of 
the  rivets  is  f  the  tensile  strength  of  the  plate.     What  is  the 
pitch? 

Ans.     6.24  ins. 

10.  In  a  pin  joint  (Fig.  6)  the  shearing  stress  of  the  pin 
is  |  the  tensile  stress  of  the  rod.     Compare  the  diameters. 

Ans.     Nearly  equal. 

11.  A  square  bar  of  steel  is  under  a  tensile  stress  of  4  tons 
and  a  compressive  stress  of  2  tons  at  right  angles  to  its  axis. 
What  are  the  shearing  and  normal  stresses  on  a  plane  making 

1 

the  angle  tan"1 — =with  the  axis? 
V2 

Ans.     S.  =  2\/2  tons.     N  =  0. 

12.  What  should  be  the  pitch  of  1.25-in.  rivets  in  a  treble- 
riveted,  double  butt  strap  joint  between  1-in.  plates,  if  the 
resistance  to  shearing  is  f  the  resistance  to  tearing? 

Ans.     6.77  ins. 

13.  What  pressure  in  a  copper  pipe,  T45  in.  in  diameter 
and  .02  in.  thick,  will  stress  the  copper  to  8000  Ibs.  per  sq.  in.? 

Ans.     800  Ibs. 


CHAPTER   III. 
TORSION. 

15.  In  Chapters  I  and  II  we  have  seen  the  effects  of 
tension  or  compression,  and  of  shearing.  In  this  chapter  we 
will  fix  firmly  one  end  of  our  rod  and  twist  it  by  applying  a 
turning  moment  to  the  free  end.  This  will  subject  it  to  tor- 
sion, with  the  effect  that  any  right  section  will  be  turned 
about  its  center  by  an  amount  depending  upon  the  distance 
of  the  section  from  the  fixed  end.*  The  stress  between  the 
particles  on  either  side  of  any  section  will  be  parallel  to  the 
section,  or  it  is  tangential  stress.  Clearly,  then,  torsion  is  a 

kind  of  shear.  If  before  ap- 
plying our  turning  moments  we 
draw  a  line,  AB  (Fig.  13),  on 
our  test  piece,  parallel  to  its 
Fio.  13  axis,  it  will,  after  the  moment 

is  applied,  be  found  to  have 
deformed  into  helix  or  screw  thread,  and  the  point  B  will 
now  be  at  6.  The  angle  BAb,  or  <£,  is  the  angle  of  torsion, 
and  is  proportional  to  the  radius  of  the  rod.  The  angle  at 
the  center,  6,  is  called  the  angle  of  twist,  and  is  proportional 
to  the  length  of  the  rod.  The  distance  Bb  is  equal  to  rd, 
but  it  is  also  equal  to  l(f>;  as  (j>  is  a  very  small  angle,  so  that 
AB  and  Ab  are  practically  equal, 

rO  =  ty;  or  <£  =  ~  .  (1) 

*  We  might  have  applied  a  turning  moment  to  both  ends  in  oppo- 
site directions,  in  which  case  the  section  at  the  middle  of  the  rod  (the 
fixed  end  as  taken  above)  would  have  remained  stationary,  though  it 
would  have  been  stressed  as  the  others. 

26 


TORSION.  27 

We  have,  by  experiment,  the  equation  q  =  Ccf)  for  torsion, 
where  q  is  the  stress  at  the  surface  and  C  is  the  modulus  of 
distortion  or  torsion,  just  as  we  have  p  =  Ee  to  hold  in  ten- 
sion or  compression,  <£,  the  angle  of  torsion,  being  a  meas- 
ure of  the  strain.  We  have  then  two  values  of  </>, 

M'^<  -I'/'  7  -•§••«'-£•        (2) 

It  is  obvious  that  for  any  material,  d,  on  any  section,  will  be 
constant  for  any  stress.  This  being  true,  our  equation  for 
6  proves  that  q  varies  with  r,  or  if  q,  ql}  q2  be  the  stress  on 
the  surface  rods  of  the  same  length  of  radii  r,  rl7  r2,  then 

?=*=?>.  (3) 

r      r,      r2 

This  is  also  true  by  experiment,  for  radii  drawn  on  any 
section  remain  straight  lines  after  torsion. 

16,  Let  us  find  the  stress  in  a  tube  so  thin  that  the 
intensity  of  the  stress  due  to  torsion  will  be  practically  the 
same  over  the  whole  sectional  area.  If  we  call  this  inten- 
sity of  stress  q,  the  mean  radius  of  the  tube  r,  and  t  the 
thickness,  then  the  area  of  any  section  will  be  2  nrt,  and 
the  stress  on  this  area  will  be  q.2  nrt,  and  the  moment  of 
this  stress  about  the  center  of  the  section  will  be  q.  2  nrt.r. 
This  must  balance  the  moment  of  the  force  applied  to 
twist  the  tube,  and,  calling  this  moment  T, 

(D 


Now  putting  this  volume  of  q  in  equation  (2)  of  Art.   15, 
which  is  true  for  either  hollow  tubes  or  solid  rods,  we  have 

IT 


for  the  angle  of  twist  of  a  thin  tube.      (Values  of  C  at  end 
of  Chapter  I.) 


28 


STRENGTH  OF   MATERIAL. 


17,   Let  us  now  take  a  tube  whose  thickness    must   be 
considered.       Let   r,   be   its   external   and    r2   its    internal 

radius  (Fig.  14),  and  let  q  equal 
the  stress  at  a  distance  r  from 
the  center  of  the  tube.  Call  ql  the 
maximum  stress  which  is  at  the 
surface,  then  from  equation  (3), 
Art.  15, 


Fig.  14.  ri  7i 

The  element  of  area  at  a  distance  r  from  the  center  is 

2  xrdr,  the  stress  on  it  is  q. 2  nrdr,  but  from  above  q  =  —  > 

ri 
so  in  terms  of  the  maximum  stress  the  stress  on  the  element 

is  —  .2  xrdr  and  the  moment  of  this  stress  is  —  .2  7ir2dr  or 
r,  r, 

-^-r3dr,  which  if  integrated  between  the  limits  of  r,  which 


are  r2  and  r,,  will   give  us  the  moment  of  the  resistance 
offered  by  the  tube  to  the  twisting  moment  applied,  or 


4  rt 


Knowing  T  this  gives  us  q^  = 


-  r2<) 


If  now  r2  =  0;    or  in  other  words,  if  the  tube  is  solid, 

2  T 


18.  It  is  clear  that  if  we  give  to  q{  the  limiting  value  for 
the  material,  we  can  find  the  diameter  of  the  solid  rod  that 
will  carry  a  given  twisting  moment,  This  formula  is  used  to 


TORSION.  .     29 

design  shafts  to  transmit  a  given  horsepower.  Let  a  be  the 
length  of  the  crank  arm,  and  F  the  mean  force  acting  on  it 
during  a  revolution;  then  the  mean  twisting  moment  is 
equal  to  aF,  and  the  work  done  per  revolution  is  equal  to 
aF.2  TT  (care  must  be  taken  to  use  the  same  units  through- 
out). The  energy  of  the  machine  per  revolution  is 

H.  P.  X  33,000 

—  foot-pounds,  where  H.  P.  is  the  indicated 

N 
horsepower  and  N  the  number  of  revolutions  per  minute. 

This  is  also  the  work  done  per  revolution,  and  if  divided  by 
2  n  will  give  the  mean  twisting  moment.  The  mean  twist- 
ing moment  is  less  than  the  maximum  twisting  moment, 
but  the  shaft  must  be  designed  to  carry  the  greatest,  which 
is  equal  to  a  constant  times  the  mean,  or  T  =  KTm'  We 

put  the  greatest  moment  equal  to  -  »  substitute  the  limit- 

ing stress  allowed  for  q  and  solve  for  r.  We  can  get  the 
greatest  twisting  moment  if  we  take  the  force  acting  on  the 
piston,  through  the  piston  and  connecting  rods  to  the  end 
of  the  crank,  when  the  crank  is  perpendicular  to  the  connect- 
ing rod,  and  multiply  this  by  the  length  of  the  crank,  remem- 
bering that  the  force  acting  along  the  connecting  rod  is  the 
force  acting  on  the  piston  multiplied  by  the  secant  of  the 
angle  the  piston  rod  makes  with  the  connecting  rod. 

19.  To  compare  solid  and  hollow  shafts,  let  the  radii  of 
the  hollow  shaft  be  r^  and  r2,  and  that  of  the  solid  shaft  be 
r;  then  if  T  is  the  resistance  offered  by  the  solid  shaft  and 
Tl  that  by  the  hollow  shaft, 

*q  (r  <      r  <) 
T       2  r      '         2         r  4     r  * 

•*•  1  _  ^  r\  _  rl    ~  "2 


xqr3  r/ 


2 

which  gives  us  the  ratio  for  shafts  of  the  same  material. 


30  STRENGTH  OF   MATERIAL. 

Now  if  the  sectional  areas  are  the  same  nr2  =  n  (r2  —  r22) 
or  r  =  Vrf  —  r22,  substituting 


Il 

T 


•which  shows  that  the  nearer  r,  approaches  rn  or  the 
thinner  the  metal  of  the  hollow  shaft  becomes,  always 
retaining  the  same  sectional  area  as  the  solid,  the  nearer  the 
ratio  of  strength  of  hollow  to  strength  of  solid  would 
approach  GO.  There  is,  however,  a  limit  to  the  thinness  of 
our  shaft,  as  it  must  be  able  to  support  its  own  weight  with- 
out buckling.  If  we  assume  that  the  ratio  of  the  radii  of 
the  hollow  shaft  is  as  2  to  1,  and  that  the  sectional  areas 

T 

are  equal,  we   get—  =  1.44,  or  the   hollow  shaft  is  nearly 

half  again  as  strong  as  the  solid  one  under  these  conditions. 
20.    Substitute  in  equation  (2)  of  Art.  15  the  value  of  q 
found  in  Art.  17  for  the  solid  shaft,  and  we  get  the  angle  of 
twist 

2  Tl 


0  = 


for  a  solid  shaft.     If  we  use  q  found  in  that  article,  for  the 
hollow  shaft  we  get 

2TI 


0 


for  the  hollow  shaft.  Now  if  T  is  known  and  6  can  be  meas- 
ured, we  can  find  the  value  of  C;  or,  if  C  is  known  and  0 
measured,  we  can  find  T,  and  thence  the  horsepower  that  a 
rotating  shaft  is  transmitting. 


TORSION.  31 

Examples  : 

1.  What  must  be  the  diameter  of  a  shaft  to  transmit  a 
twisting  moment  of  352  ton-ins.,  the  stress   allowed   being 
3£  tons  per  sq.  in.?     What  H.   P.  would  this  shaft  trans- 
mit at  108  revolutions  per  minute,  the  maximum  twisting 
moment  being  H  of  the  mean? 

Solution: 

2T  2T     2X352X7X2 

q  = or  r3  = =  -      — — =  64. 

Trr3  nq  22  X  7 

.'.r  =  4  ins.;  d  =  8  ins. 

_55     H.  P.  X  33,000  X  12 
~36  '       N  X  In  X  2240 

352  X  2240  X  108  X  2  X  22  X  36 
33,000  X  12  X  7  X  55 

2.  Find  the  size  of  a  hollow  shaft  to  replace  the  preced- 
ing, exterior  diameter  to  be  I   the  interior.     What  is  the 
weight  of  metal  saved  in  a  steel  shaft  60  ft.  long?     (Cu.  ft. 
steel  =  480  Ibs.) 

Ans.     Exterior    diameter    8.454    ins.     Interior    diameter 
5.284.     Weight  saved  3213  Ibs. 

3.  Find  the  diameter  of  a  solid  shaft  to  transmit  9000 
H.  P.    at  140   revolutions  per   minute,    the  stress   allowed 
being  10,000  Ibs.  per  sq.  in.,  and  the  maximum  twisting 
moment  I  the  mean. 

Ans.     d  =  14.5695  ins. 

4.  Find  the  size  of  a  hollow  steel  shaft  to  replace  the  above, 
internal  diameter  being  T9s  of  the  external.     What  is  the 
saving  in  weight  for  60  ft.  of  shafting? 

Ans.     External  diameter  =  15.091  ins.     Internal  diameter 
=  8.4886  ins.     Weight  saved  8893.5  Ibs, 


32  STRENGTH  OF  MATERIAL. 

5.  Compare  the  strength  of  a  solid  wrought-iron  shaft  with 
a  hollow  steel  shaft  of  the  same  external  diameter,  the  in- 
ternal diameter  of  the  steel  shaft  being  \  the  external  and 
the  elastic  strength  of  steel  §  that  of  iron. 

Ans.     As  32  is  to  45. 

6.  A  solid  shaft  fits  exactly  inside  a  hollow  shaft  of  equal 
length.     They  contain  the  same  amount  of  material.     Com- 
pare their  strengths  when  used  separately. 

Ans.     As  3  is  to  \/2. 

7.  The  resistance  of  a  twin-screw  vessel  at  18  knots  is 
44,000  Ibs.     At  95  revolutions  per  minute  what  will  be  the 
twisting  moment  on  each  shaft?     What  is  the  H.  P.? 

Ans.     T  =  360  in.-tons.     H.  P.  =  2432. 

8.  The  pitch  of  a  screw  propeller  is   14  ft.;  the  twist- 
ing moment   on   the   shaft   is   120   ton-ins.;  the  mean  dia- 
meter of  the  thrust  bearing  rings  is  15  ins.;  coefficient  of 
friction  is  .05;  find  the  thrust  and  the  efficiency  of  the  thrust 
bearing. 

Ans.     Thrust  =  4^  tons.     Efficiency  of  bearing  98t%. 

9.  The  angle  of  torsion  of  a  shaft  is  not  to  exceed  1°  for 
each  10  ft.  of  length;  what  must  be  its  diameter  for  a  twisting 
moment  of  16,940  ft.-lbs.? 

f  221 

(7=10,976,000  in.-lb.  units.     *  =  — I. 

Ans.     Diameter  =  6  ins. 

10.  A  solid  steel  shaft,  10.63  ins.  in  diameter,  is  transmit- 
ting 12,000  H.  P.   at  200  revolutions  per  minute.     If  the 
maximum  twisting  moment  is  I  the  mean,  what  is  the  maxi- 
mum stress  (torsion)  in  the  shaft  ? 

Ans.     20,000  Ibs.  per  sq.  in. 


TORSION.  33 

11.  If  the  modulus  of  rigidity  be  4800  in  in.-ton  units, 
what  is  the  greatest  stress  to  which  a  shaft  should  be  sub- 
jected that  the  angle  of  torsion  may  not  exceed  1°  for  each 
10  diameters  of  length? 

Ans.     4.2  tons  per  sq.  in. 

12.  In  changing  engines  in  a  ship,  the  number  of  revolu- 
tions is  increased  J;  H.  P.  is  doubled;  the  ratio  of  maximum 
to  mean  twisting  moment  is  changed  from  f  to  £;  and  the 
strength  of  material  of  the  shaft  is  25%  greater.     What  is  the 
relative  size  of  the  new  shaft? 

Ans.     The  same  size. 

13.  Find  the  angle  of  torsion  of  a  steel  tube,  6  ft.  long, 
J  in.  thick;  mean  diameter  12  ins.,  shearing  stress  allowed 
4  tons  per  sq.  in. 

Ans.     0°.0478, 


CHAPTER  IV. 
BENDING. 

21.  There  is  one  more  way  in  which  we  can  introduce  stress 
into  our  rod  and  that  is  by  bending  it.  If  we  rest  our  rod  on 
supports  at  its  ends  and  load  it  at  its  middle,  it  will  bend  into 
a  curve.  The  plane  of  this  curve  is  called  the  plane  of  bend- 
ing (OABO,  Fig.  15).  The  particles  on  the  concave  side  of 


Fig.  15. 

the  rod  will  tend  to  crowd  together  and  will  be  in  compres- 
sion, while  those  on  the  convex  side  will  tend  to  pull  apart  and 
will  be  in  tension;  obviously,  there  will  be  some  intermediate 
plane,  perpendicular  to  the  plane  of  bending,  where  there  will 
be  neither  tension  nor  compression,  a  plane  of  no  stress.  This 
plane  is  called  the  neutral  surface.  The  neutral  line  is  the 
intersection  of  the  neutral  surface  with  the  plane  of  bending, 
and  it  gives  us  the  curve  into  which  the  rod  bends,  and  is 
therefore  called  the  elastic  curve.  The  intersection  of  the 
neutral  surface  with  any  section  of  our  rod  perpendicular  to 
its  axis  is  called  the  neutral  axis  of  that  section.  In  Fig.  15, 
the  lines  AO  and  BO  meet,  0  being  their  point  of  intersec- 

34 


BENDING. 


35 


tion.  AOB  is  the  plane  of  bending.  CEFD  is  the  neutral 
surface.  A  B  is  the  neutral  line  or  elastic  curve;  and  if  HH  is 
any  section  of  the  rod  perpendicular  to  AB,  then  GK  is  its 
neutral  axis. 

22.  Now  before  bending,  all  sections  of  our  rod  are  parallel, 
but  after  bending  these  same  sections  (assuming  that  they 
remain  plane)  will  be  inclined  to  each  other,  being  nearer 
together  on  the  concave  side, 
the  same  distance  apart  as 
before  bending  at  the  neutral 
surface,  and  farther  apart  at 
the  convex  side.  If  R  is  the 
radius  of  curvature  of  the 
neutral  line  AB  (Fig.  16),  6 
the  angle  between  the  planes 
of  any  two  sections,  and 
y  the  distance  from  the 
neutral  plane  to  any  point 
in  the  rod,  then  the  length 
of  the  elastic  curve  between 


Fig.  16. 


these  two  sections  is  R6,  and  the  length  between  these  two 
sections,  along  a  line  all  points  of  which  are  at  a  distance  y 
from  the  neutral  surface  is  (R  ±  y)  0,  or  Rd  ±  yd,  so  that 
yd  is  the  total  change  of  length  between  the  two  sections  at  a 
distance  y  from  the  neutral  surface.  The  change  of  length 
per  unit  length  is  the  total  change  divided  by  the  original 

length,  or  the  strain  is  —  =  —  •     From  Chapter  I,  Art.  5, 

Rd     R 

P 

we  have  the  strain  due  to  tension  or  compression  equal  to  — . 

E 

p    y 

Hence  —=7-,  or  the  stress  due  to  bending  at  a  distance  y 
hi      R 

from  the  neutral  plane  will  be 

*  E 


36  STRENGTH  OF  MATERIAL. 

The  stress  varies  then  as  the  distance  from  the  neutral  surface, 
and  must  not  at  the  outer  surface  in  the  plane  of  bending 
exceed  the  elastic  limit  of  the  material,  or  the  surface  fiber 
stress  must  be  less  than  /. 

23.  We  have  above  our  formula  for  the  fiber  stress,  but  do 
not  know  what  value  to  use  for  y,  as  where  the'  neutral  axis 
lies  is  not  known.  Neither  do  we  know  the  length  of  R.  We 
will  first  prove  that  the  neutral  axis  of  any  section  of  a  beam 
passes  through  the  center  of  gravity  of  the  section.  We  know 
that  the  stress  is  maximum  compressive  at  the  surface  on  the 
concave  side;  decreases  with  the  distance  from  the  neutral 
surface,  where  it  is  zero;  there  changes  to  tension;  and  in- 
creases to  a  maximum  at  the  surface  on  the  convex  side.  The 
rod  is  in  equilibrium,  a  law  of  which  is  that  the  sum  of  the 
horizontal  components  of  all  the  forces  acting  must  equal 
zero.  Now  the  loads  which  cause  the  bending  are  all  vertical, 

so  can  have  no  horizontal 
components;  the  only  other 
forces  are  the  horizontal 
stresses  at  the  section,  and 
their  sum  must  be  equal  to 
17  zero,  the  stresses  on  the 

opposite  sides  being  equal 
but  opposite  in  direction.  Fig.  17  shows  the  stress  on  one 
side  only,  and  the  section  may  be  of  any  shape  whatever. 

E 

Let  AB  be  the  neutral  axis.     The  formula  p  =  —  y  gives  us 

R 

the  stress  per  unit  area  on  any  element  of  area,  dA,  which  is 
throughout  at  a  distance  y  from  the  neutral  axis.  The  stress 

E 

on  the  element  is  then  pdA,  or  —  yd  A.     If  we  integrate  this 

R 

between  the  limits  for  our  section,  we  will  get  the  total  stress 
on  the  section,  which  we  know  to  be  equal  to  zero,  or 

E  r11"*1' 

H  =  -JydA=0.  (1) 


** 


limit 


BENDING.  37 

Now  the  value  of  the  integral,  f yd  A,  divided  by  the  area  of 
the  section  will  give  us  the  distance  of  the  center  of  gravity 
of  the  section  from  the  line  AB,  and  if  this  distance  were 
zero  the  line  would  pass  through  the  center  of  gravity  of  the 
section.  Now  equation  (1)  equals  0;  we  know  E  and  R  have 

/limit 
ydA  must  be  equal  to 

limit 

zero.  Hence  the  neutral  axis  always  passes  through  the 
center  of  gravity  of  the  section. 

24.  Determination  of  R.  —  Our  rod  is  in  equilibrium,  one 
of  the  laws  of  which  is  that  the  sum  of  the  moments  of  all  the 
forces  about  any  axis  must  equal  zero.  Being  true  for  any 
axis  we  will  take  the  neutral  ,w 

axis  for  the  axis  of  moments. 
We  will  first  find  the  sum  of 
the  moments  of  the  external 
forces.  In  Fig.  18  let  AB  be 
the  rod  of  length  /,  loaded  with 
W  pounds  in  the  middle,  so  that  the  supporting  forces  are 

W 
each  — ;  then  to  the  left  of  the  section  HH  the  moment  of 

W      W 
the  supporting  force  —  is  — x;  and  the  moment  of  the  forces 

to  the  right  is 

Wx 

T' 

the  same  value  as  before.  The  moment  to  the  left  tends  to 
turn  the  part  of  the  beam  on  the  left  of  the  section  in  the 
direction  of  motion  of  the  hands  of  a  watch,  while  that  to  the 
right  tends  to  turn  the  right  part  of  the  beam  in  the  opposite 
direction;  therefore,  if  we  call  the  first  direction  positive,  the 
other  is  negative.  As  in  the  other  cases  of  equilibrium  we 
will  use  the  values  found  on  one  side  only  of  the  section.  In 


38  STRENGTH  OF  MATERIAL. 

this  case,  the  moment  of  the  external  forces  about  the  neutral 
axis  is  the  bending  moment  for  that  section,  and  we  will 

Wx 
designate  it  by  M,  so  M  =••  --     Considering  then  the  part 

to  the  left  of  this  section,  the  moment  of  the  stress  in  the  sec- 
tion must  balance  the  moment  of  the  external  forces  about  the 
neutral  axis.  Fig.  19  shows  enlarged  the  part  of  the  rod  to 

the  left  of  the  section.  We 
see  that  the  moment  of  the 
external  forces  tends  to  turn 
this  part  of  the  rod  in  the 
direction  indicated  by  the 
arrow  marked  (1),  while 
the  moment  of  the  stress  in 
Fig.  19.  the  section  tends  to  turn 

it  in   the   direction  of   the 
arrow  (2),  and  for  equilibrium  these  moments  must  be  equal. 

El 

As  in  Art.  23,  the  stress  on  the  area  dA  is  —  yd  A,  and  the 

it 

Tjl 

moment  is  —ydA.y,  which  integrated  between  the  limits  for 
H 


ft      /*imit 

the  sections  gives—  I   y2dA.     The  integral  f\fd  A  between 

R  «/limit 

limits  is  the  moment  of  inertia  of  the  area  of  the  section 
about  the-  neutral  axis,  and  we  will  designate  it  by  /,  so  that 

E 

the  moment  of  the  stress  in  the  section  is  equal  to  —  /.    This 

R 

W  V 

must  equal  M  ,  so  M  -•=  —  /,  or  R  =  —  I.     In  Art.  22  we  found 

E     p 

—  =  _,  so  we  now  have  a  general  formula  for  bending, 
R     y 

PME 


BENDING.  39 

In  which     p  —  stress  at  any  point  at  a  distance  y  from  the 

neutral  surface; 

M  =  bending  moment  at  any  section  due  to  exter- 
nal loads; 
/  =  moment  of  inertia  of  the  area  of  the  section 

about  the  neutral  axis; 

R  =  radius  of  the  arc  into  which  the  rod  is  bent; 
E  =  modulus  of  elasticity  of  the  material  of  the  rod. 

25.  We  have  now  seen  the  effect  of  applying  forces  to  our 
rod  in  all  the  different  ways.  If  the  stresses  caused  have  the 
same  line  of  action,  that  is,  if  they  act  on  any  section  in  the 
same  or 'in  a  diametrically  opposite  direction,  their  algebraic 
sum  will  give  us  the  total  stress  on  the  section.  For  example, 
the  stress  due  to  several  longitudinal  loads  is  the  algebraic 
sum  of  the  loads  divided  by  the  area  of  the  section;  or,  if  a 
rod  is  under  a  tensile  stress  and  there  is  also  a  fiber  stress  due 
to  bending,  the  maximum  stress  would  be  at  the  convex  sur- 
face and  equal  to  the  sum  of  the  two  stresses,  while  at  the  con- 
cave surface,  which  would  be  in  compression  due  to  the 
bending,  the  stress  would  be  the  difference  between  the  two, 
and  would  act  in  the  direction  of  the  greater.  Another  fact 
must  be  considered  in  actual  practice :  When  the  temperature 
changes,  a  free  rod  expands  or  contracts  without  stress,  but 
if  the  rod  be  prevented  from  expanding  or  contracting,  stress 
is  produced.  The  method  of  taking  the  algebraic  sum  when 
the  lines  of  action  are  the  same  is  called  the  principle  of 
superposition,  which  may  be  stated  as  follows:  The  effect 
due  to  a  combination  of  forces  is  equal  to  the  sum  of  the 
effects  due  to  each  force  taken  separately. 

When  the  stresses  caused  by  our  forces  have  not  the  same 
line  of  action,  such  as  combinations  of  shearing  or  torsion 
with  bending,  we  must  arrive  at  their  maximum  effects  in 
some  other  way.  In  the  next  chapter  we  will  endeavor  to 
show  how  to  find  the  maximum  stress  together  with  its  direc- 


40  STRENGTH  OF  MATERIAL. 

tion,  which  is  due  to  the  combined  effects  of  two  or  more 
stresses  which  act  at  right  angles.  We  will  then  be  able  to 
combine  the  effects  of  any  of  the  forces  which  we  have  applied 
separately  to  our  rod  in  these  first  four  chapters.  We  will 
find  that  the  maximum  stresses  due  to  any  of  these  combina- 
tions are  greater  than  any  of  the  stresses  acting  singly,  and 
that  they  act  on  planes  which  are  inclined  to  those  on  which 
any  of  the  single  stresses  act.  This  will  account  for  the 
apparently  erratic  manner  in  which  material  sometimes  car- 
ries away;  so  in  deciding  if  a  single  part  of  a  machine  or 
structure  is  strong  enough,  we  must  first  find  to  what 
forces  it  is  subjected,  and  then  if  it  be  able  to  sustain  the 

total  stresses  they  induce. 

I 

Examples : 

1.  A  beam  2  ins.  wide  by  3  ins.  deep  is  subjected  to  a 
bending  moment  of  72  ton-ins.     What  is  the  maximum  fiber 
stress? 

Solution:        p       M  My  3 

y--j      P--—     y=2- 

1  9  72  X  3  X  2 

I=i2Ah  =2-  •'•p-  ~2irr  =24tons- 

2.  An  iron  I-beam  (without  weight)  of  12-ft.  span  has 
flanges  4  ins.  by  1  in.,  and  web  8  ins.  by  \  in.  *  What  is  the 
greatest  central  load  it  can  carry  if  the  stress  is  limited  to 
4  tons  per  sq.  in.? 

Solution  : 

••  y.    Mmax  =         =  W  X  3  ft.-lbs.  =W  X  36  in.-lbs., 

_  552  4_TFX36X3 

l=—.y-  5.       -  = 


=  = 

5X36X3 


BENDING.  41 

3.  A  cast-iron  I-beam  has  a  top  flange  3  ins.  by  1  in. ;  bot- 
tom flange  8  ins.  by  2  ins.;  web,  trapezoidal,  J  in.  thick  at 
top  and  1  in.  thick  at  bottom;  total  depth  of  beam  16  ins. 
Find  the  position  of  the  neutral  axis  and  the  ratio  of  maxi- 
mum tensile  to  compressive  stresses. 

Ans.  Neutral  axis  4.81  ins.  from  bottom.  Ratio  T  to  C 
is  3  to  7. 

•  4.  A  wooden  beam  of  rectangular  cross-section  is  15  ft. 
long  and  10  ins.  wide.  If  the  maximum  bending  moment  is 
16.5  ton-ft.,  and  the  allowed  stress  is  \  ton  per  sq.  in.,  what  is 
its  depth? 

Ans.     15.4  ins. 

5.  An  I-beam  is  25  ft.  long,  top  flange  3  ins.  by  2  ins.; 
bottom  flange  10  ins.  by  3  ins.;  web  12  ins.  by  1  in.;  total 
depth  17  ins.     If  the  stress  is  limited  to  4J  tons  per  sq.  in., 
find  the  greatest  central  load  it  can  support  in  addition  to 
its  own  weight  (take  weight  of  beam  as  2000  Ibs.  acting  at 
its  center). 

Ans.     6.48  tons. 

6.  What  is  the  radius  of  the  smallest  circle  into  which  a 
rod  of  iron  2  ins.  in  diameter  may  be  bent  without  injury,  the 
stress  being  limited  to  4  tons  per  sq.  in.     E  =  13,000  ton-ins. 

Ans.     R  =  270  ft.  10  ins. 

7.  A  spar,  20  ft.  long,  is  supported  at  the  ends  and  sus- 
tains a  maximum  bending  moment  of  3147.5  Ib.-ft.     If  the 
stress  be  limited  to  J  ton  per  sq.  in.,  what  is  the  diameter  of 
the  spar? 

Ans.     7.0025  ins. 

8.  What  is  the  diameter  of  the  smallest  circle  into  which  a 
^-in.  steel  wire  may  be  coiled,  keeping  the  stress  within  6  tons 
per  sq.  in.?  (E  for  steel  wire  being  35,840,000  in  in.-lb.  units.) 

Ans.     111J  ft. 


42  STRENGTH  OF  MATERIAL. 

9.  A  rectangular  beam  12  ft.  long,  3  ins.  wide,  9  ins.  deep, 
is  supported  at  the  ends.     Stress  is  limited  to  3  tons  per  sq. 
in.     Find  the  load  which  can  be  carried  at  the  center;  also 
find  the  load  if  the  beam  lies  the  flat  way,  i.e.,  3  ins.  deep  and 
9  ins.  wide. 

Ans.     1st  case  3f  tons;  2d  case  1J  tons. 

10.  Find  the  breadth  and  depth  of  the  rectangular  beam  of 
maximum  strength  which  can  be  sawed  from  a  log  2  ft.  in 
diameter,  and  compare  its  resistance  to  bending  with  that  of 
the  largest  square  beam  that  can  be  sawed  from  the  same  log. 

2  2\/2 

Ans.     Top  is  — -ft.,  depth  is — =-  ft.     Resist  bending  in 

V3  A/3 

ratio  of  1.089  to  1. 

11.  A  steel  I-beam  30  ft.  long;  flanges  7  ins.  by  .8  in.; 
web  24  ins.  by  .5  in.;  carries  31,900  Ibs.  at  its  center.     What 
is  the  maximum  fiber  stress? 

Ans.     16,000  Ibs.  per  sq.  in. 

12.  Compare  the  resistance  to  bending  of  a  wrought-iron 
I-beam;  flanges  6  ins.  by  1  in.;  web  8  ins.  by  J  in.,  when  up- 
right, and  when  laid  on  its  side. 

Ans.     4.6  to  1. 

13.  A  round  steel  rod,  2  ins.  in  diameter,  can  only  with- 
stand a  bending  moment  of  6  ton-ins.     What  is  the  greatest 
length  of  such  a  rod  which  will  just  carry  its  own  weight  when 
supported  at  the  ends? 

Ans.     28J  ft. 


CHAPTER   V. 

COMBINATION  OF  STRESSES. 

26.  In  a  rod  suffering  tension  and  shear  let  HH  (Fig.  20) 
be  any  section  which  has  normal  stress  due  to  the  tensile  load, 
F,  and  tangential  stress  due  to  shear.  Let  the  square  shown 
represent  the  base  of  an  elementary  cube  of  volume,  whose 
height  dz  is  perpendicular  to  the  plane  of  the  paper,  and  on 
whose  faces  dydz  are  the  stresses  P  and  S  as  shown.  It  is 


H 

\ 

—  >F 

PJ[ 

dot 

fr 

t 
Fig. 

H 

20. 

clear  that  the  stresses  P  balance  each  other,  but  it  will  be 
noticed  that  the  tangential  stresses  S  form  a  moment  whose 
effort  is  to  turn  the  cube  to  the  right.  Now  the  forces  are 
assumed  within  the  elastic  limit  and  we  know  the  cube  to  be  in 
equilibrium,  so  there  must  be  an  equal  and  opposite  moment 
tending  to  turn  it  to  the  left.  In  other  words,  we  must  have 
tangential  stresses  equal  to  S  on  the  faces  dxdz,  acting  in  a 
direction  such  that  their  moment  will  turn  the  cube  to  the 
left.  These  latter  stresses  are  called  longitudinal  shear,  and 
because  of  the  resistance  offered  to  it  a  solid  beam  will  bend 
less,  when  supported  at  the  ends,  than  a  pile  of  thin  boards  of 
the  same  volume.  We  may  then  concede  that  the  shearing 
stresses  at  any  point  within  a  body  in  a  state  of  stress  are 
equal  and  act  in  planes  at  right  angles  with  each  other. 

43 


44 


STRENGTH   OF   MATERIAL. 


27.  We  can  now  find  the  plane  on  which  the  resultant 
stress,  due  to  a  number  of  stresses  acting  in  planes  at  right 
angles,  is  a  maximum.  Fig,  21  is,  enlarged,  the  elementary 
cube  of  Fig.  20,  with  the  additional  stress  Plf  acting  so  as  to 

put    it    in    vertical    tension. 
-&    Let  AB  be  any  plane  making 
the  angle  a  with  the  direction 
*-P    of  the  stress  P,   and   let  the 
sum  of  the  components  of  all 
the  stresses    on    one   side   of 
this     plane,     when    resolved 

—____>  c  w 

|  along    and    perpendicular    to 

pi  it,    be    N    and    T  as   shown. 

The  stresses  on  the  other  side 

of  A B  would  give  equal  and  opposite  components  to  N  and 
T  (equilibrium).  So  that  our  cube  is  now  in  equilibrium 
under  the  stresses  shown.  Let  us  resolve  these  stresses 
horizontally  and  vertically,  then  the  sum  of  both  the  hori- 
zontal and  the  vertical  components  will  be  zero.  For 
example,  the  intensity  of  the  stress  P  on  AB  is  P  sin  a, 
remembering  that  the  area  of  the  inclined  plane  is  the  area 
of  face  dydz  X  cosec  a.  (See  Art.  14.)  Resolving  then 
we  have 

N  sin  a  +  T  cos  a  —  P  sin  a  —  S  cos  a.  =  0  (1)  horizontally. 
N  cos  a  —  T  sin  a  —  Pl  cos  a  —  S  sin  a  =  0  (2)  vertically. 

Eliminating  T  between  (1)  and  (2)  we  get 

N  =  P  sin2  a  +  Pl  cos2  a  +  2  S  sin  a  cos  a.  (3) 

Reducing 


(sin2  a  = 


1  —  cos  2  a 


a  — 


1  +  cos  2  a 


and  2  sin  a  cos  a  =  sin  2  a  j 
cos  2  a  +  S  sin  2  a.  (4) 


COMBINATION  OF  STRESSES.  45 


By  the  same  method  we  get 


P  -  Pi 

T  =  -         -  sin  2  a  +  S  cos  2  a.  (5) 


(4)  and  (5)  give  us  the  stresses  along  and  perpendicular 
to  any  plane  due  to  the  combined  stresses  P,  Pl  and  S. 
Putting  equal  to  zero  the  first  derivative  with  regard  to  N 
and  a  or  T  and  a  in  (4)  and  (5)  will  give  us  the  value  of  a 
for  which  N  or  T  is  a  maximum  or  minimum. 

—  =  o  =  --  -  --  2  sin  2  a  +  2  S  cos  2  a 
da  2 

gives 

tan  2  a  =  JL         or  a  =  J  tan"1     A-+      .       (6) 


7/T7  P  p 

—  =  0  =  -  -  2  cos  2  a  -  2  S  sin  2  a 
da  2 

gives 

p    _     p  D    _     E)    i-    .      _ 

tan  2  a  =  -^-^  or  a  =  i  tan'1  -^-i^+  _  .          (7) 

28.  Equation  (6)  gives  us  the  angle  with  P  of  a  plane  on 
which  the  normal  stress  due  to  the  combined  load  is  a  maxi- 
mum (the  plane  90°  from  it  giving  the  minimum),  and 
equation  (7)  the  angle  with  P  of  a  plane  on  which  the 
tangential  stress  due  to  the  combined  load  is  a  maximum 
(minimum  90°  from  it).  Obviously,  from  these  equations, 
the  two  planes  are  at  45°  from  each  other  (tan  2  a  being  in 
one  case  the  negative  reciprocal  of  what  it  is  in  the  other), 
or  the  planes  of  maximum  normal  and  maximum  tangential 
stresses  lie  at  angles'  of  45°  with  each  other.  Equation  (6) 
shows  the  maximum  and  minimum  normal  stresses  to  be  at 
right  angles,  and  if  we  substitute  the  value  of  2  a  found 
from  it,  in  equation  (5)  we  find  that  on  this  plane  of  maxi- 


46  STRENGTH   OF  MATERIAL. 

mum  normal  stress  the  tangential  stress  is  zero,  or  the  shear 
is  zero  when  the  normal  stress  is  a  maximum.  The  normal 
stresses  found  in  equation  (6)  and  acting  on  planes  at  right 
angles  are  called  principal  stresses  and  their  directions  prin- 
cipal directions.  Principal  stresses  are  always  at  right  angles. 
If  we  substitute  in  equations  (4)  and  (5)  the  values  of 
sin  2  a  and  cos  2  a,  obtained  from  equations  (6)  and  (7) 
respectively,  we  will  get  the  maximum  value  of  the  normal 
and  tangential  stresses. 


andT7- 

2 

j    v    <±  o      T 

\^r    —  r 

i)     {°J 

(9) 

-f  sigii  max. 
-  sign  min. 

±  i  V4S 

2  +  (P  -  , 

Pi)2 

The  +  sign  of  (8)  gives  the  maximum  tension  (or  compres- 
sion), and  the  —  sign  gives  the  maximum  compression  (or 
tension)  on  a  plane  at  right  angles,  i.e.,  principal  stresses. 
Using  the  formula  of  this  article  we  can  find  the  maximum 
value  and  its  direction  of  any  combination  of  stresses  due  to 
the  loads  used  in  the  preceding  chapters. 

29.  The  Stress  Ellipse.  —  Let  us  assume  that  at  any 
point  within  a  body  there  are  two  normal  stresses  of 
intensity,  P  and  Pl  (Fig.  22),  acting  at  right  angles.  Then 
by  Art.  14,  x,  the  intensity  of  the  stress  P  on  any  plane 
making  an  angle  a.  with  the  direction  of  P,  is  P  sin  a,  and  ?/, 
the  intensity  of  the  stress  Pl  on  the  same  plane,  is  Pl  cos  a, 

x  y 

-  =  sin  a,  and  —  =  cos  a. 

Squaring  and  adding  we  have 


2 
H  --  -  =  sin2  a  +  cos2  a  =  1. 


The  above  is  the  equation  of  an  ellipse  of  which  the  semi- 
axes  are  P  and  P1;  and  of  which  x  and  y,  the  coordinates  of 


COMBINATION  OF  STRESSES. 


47 


any  point  (C,  Fig.  22),  represent  respectively  the  intensity 
of  the  stresses  P  and  Pl  on  a  plane  making  the  angle  a  with 
P.  The  radius  vector,  OC,  represents  on  the  same  scale 
the  amount  and  direction  of  the  resultant  intensity  of  stress 


,' 


Fig.  22. 


on  the  plane  AB.  The  equation  of  this  ellipse  in  terms  of 
the  eccentric  angle  <fi  is  x  =  P  cos  <£,  and  y  =  Pl  sin  <p. 
Now  the  tangent  of  the  angle  d,  which  the  resultant  stress 
makes  with  the  direction  of  P,  is 


y      P,  sin 
tan  0  =  -  = 


x       P  cos 


Pt  cos  a       Pl 

—  —      —  —tan 
P  sin  a        P 


Pl 

--  cot  a. 
P 


72,  the  resultant  stress,  is  equal  to  \/  x2  +  ?y2,  and  the 
angle  the  resultant  stress  makes  with  the  plane  AB  is 
(0  +  «)•  Had  there  been  another  normal  stress  at  right 
angles  to  the  plane  of  P  and  P0  the  locus  of  the  end  of  the 
resultant,  (7,  would  have  been  the  surface  of  an  ellipsoid. 
In  this  case  it  is  probably  as  easy  to  find  the  resultant  of 
two  of  the  stresses  and  then  get  the  resultant  of  the  third 
with  it.  This  ellipse  is  convenient  to  find  the  direction  and 
amount  of  the  resultant  of  the  principal  stresses,  or  the  com- 
binations of  other  normal  stresses,  which  may  be  done  graph* 


48 


STRENGTH  OF  MATERIAL. 


with  centers  at  0. 


ically  as  follows:  Draw  the  axes  OX  and  OY  (Fig.  23)  and 
lay  off  to  the  same  scale  P  on  OX,  and  Pl  on  OY.  With 
these  distances  for  radii,  describe  two  concentric  circles 
Draw  the  line  AB,  making  the  given 
angle  a  with  P  and  draw  OD 
perpendicular  to  it.  Where  OD 
cuts  into  the  circle  of  radius  P, 
drop  a  perpendicular  on  OX, 
and  where  it  cuts  the  circle  of 
radius  Plt  draw  FC  parallel  to 
OX.  C,  the  intersection  of  these 
last  two  lines  is  a  point  on 
the  ellipse.  CE  represents  the 
stress  due  to  P  resolved  on  the 
plane  AB,  and  OE  represents 
the  stress  due  to  Pl  resolved 
on  the  plane  AB;  while  OC 


Fig.  23. 


represents  the  resultant  stress  on  A B  due  to  both,  and  the 
angle  COB  is  the  angle  it  makes  with  the  plane  A  B. 

30.  Now  any  condition  of  stress  at  any  point  within  a 
body  may  always  be  resolved  (Art.  14)  into  three  simple 
stresses  acting  on  planes  at  right  angles.  By  means  of 
the  formula  of  Arts.  27  and  28,  we  can  always  calculate 
the  value  and  direction  of  the  resultant  stress  on  any 
plane  due  to  these  three  simple  stresses,  and  by  means 
of  the  stress  ellipse  of  Art.  29  we  can  calculate  or  deter- 
mine graphically  the  amount  and  direction  of  the  result- 
ant of  two  simple  normal  stresses  on  any  given  plane. 
For  example,  if  we  find  a  piece  of  material  under  tensile 
stress  combined  with  torsion,  we  can  substitute  for  P  in 
equation  (8)  the  tensile  stress,  and  for  S  the  stress  at  the 
surface  (maximum)  due  to  torsion;  and,  unless  there  is 
another  stress  at  right  angles  to  the  plane  of  these  two,  Pl 
is  zero.  -  The  value  of  N  given  by  these  substitutions  is  the 
maximum  normal  stress,  and  the  plane  on  which  it  acts  will 


COMBINATION  OF  STRESSES.  49 

be  found  by  making  the  same  substitutions  in  equation  (6), 
remembering  that  a  is  the  angle  the  plane  makes  with  the 
direction  of  the  tensile  stress  P.  In  case  of  bending,  the 
maximum  fiber  stress  should  be  used  for  P.  We  must 
always  use  the  maximum  values,  for  the  piece  of  material 
must  be  strong  enough  to  sustain  the  greatest  stresses  to 
which  it  will  be  subjected. 

Example  5,  at  the  end  of  this  chapter,  offers  a  good  illus- 
tration of  the  use  of  the  stress  ellipse. 

Examples : 

1.  The  shaft  of  a  vessel,  15  ins.  in  diameter,  is  subject  to 
a  twisting  moment  of  100  ft.-tons  and  a  bending  moment 
of  20  ft.-tons,  also  the  thrust  of  the  screw  is  16  tons.  Find 
the  maximum  stresses  on  the  shaft. 

Solution : 

1  "  "   2~~  f       ' 

or  P2 

N  (N  -  P)  =  S2  and  T2  =  S2  +  —  . 

4 

nd2  16       64 

A  =  --  •         /.  thrust  =  — -  =  — -  • 
4  Ttd2      nd2 


_  My  _  64  _  64 

Fa  (twisting  moment)       64  X  20 
q  (torsion)  =  S  =  - 

nd2  ?rcP 

Te 


50  STRENGTH  OF  MATERIAL. 

2.  A  tube,  12  ins.  mean  diameter  and  J  in.  thick,  is  acted 
on  by  a  thrust  of  20  tons  and  a  twisting  moment  of  25 
ft.-tons.       What    are    the    maximum    stresses?  and   their 
angles? 

Ans.     Greater  3.24  tons.     39|°. 

3.  A  rivet  is  under  shearing  stress  of  4  tons  per  sq.  in. 
and  tensile  stress,  due  to  contraction,  of  3  tons  per  sq.  in. 
What  are  the  maximum  stresses? 

Ans.     Greater  5.8  tons. 

4.  The  thrust  of  a  screw  is  20  tons;  the  shaft,  diameter  14 
ins.,  has  a  twisting   moment  of   100  ton-ft.  and  a  bending 
moment  of  25  ton-ft.     Find  the  maximum  stress  and  com- 
pare it  with  what  it  would  have  been  without  the  bending 
moment  or  thrust. 

Ans.     Greater  2.9  tons.     Ratio  1.32  to  1. 

5.  At  a  point  within  a  solid  in  a  state  of  stress  the  prin- 
cipal stresses  are  tension  of  255  and   171  Ibs.     Find  the 
amount  and  direction  of  the  stress  on  a  plane  making  an 
angle  of  27°  with  the  255  Ibs.  stress. 

Ans.     191.35  Ibs.     79°  46'  20". 

6.  A  beam  is  under  a  300-lb.  tensile  stress  and  a  100-lb. 
shearing  stress.     Find  the  normal  and  tangential  stresses  on 
a  plane  making  an  angle  of  30°  with  the  tensile  stress. 

Ans.     Normal  161:5;   tangential  179.8  Ibs.  per  sq.  in. 

7.  Find  the  principal  stresses  and  their  directions  for  a 
point  in  a  beam  which  is  under  tensile  stress  of  400  Ibs.  and 
shearing  stress  of  250  Ibs. 

Ans.     Normal    maximum    520;    minimum     —120;    X  = 
-  25°  40'  12";  or  64°  19'  48". 

8.  Find  the  maximum  and  minimum  values  of  the  shear 
in  example  7  and  their  directions. 

Ans.     320  Ibs.  per  sq.  in.     X  =  19°  19'  48";  109°  19'  48". 


COMBINATION   OF  STRESSES.  51 

0.  A  bolt,  1  in.  in  diameter,  is  under  a  tension  of  5000  Ibs. 
and  a  shearing  force  of  3000  Ibs.  Find  the  maximum  stresses 
and  their  directions. 

Ans.  N  =  8155;  T  =  4970  Ibs.  per  sq.  in.  Angles  are 
N  64°  53',  and  T  19°  53'  with  axis  of  bolt. 

10.  Find  the  maximum  stress  in  a  shaft  3  in.  in  diameter 
and  12  ft.  between  bearings,  which  transmits  40  H.  P.  at  120 
revolutions  per  minute,  and  has  a  weight  of  800  Ibs.  half  way 
between  bearings.     The  shearing  stress  due  to  above  arrange- 
ment is  4000  Ibs.  per  sq.  in. 

Ans.     N  =  7600  Ibs.  and  T  =  4900  Ibs.  per  sq.  in. 

11.  What  is  the  diameter  of  a  steel  shaft  to  transmit  90 
H.  P.  at  250  revolutions  per  minute,  the  distance  between 
bearings  to  be  8  ft.  and  a  load  of  480  Ibs.  to  be  carried  half 
way   between   bearings?     The  allowable  maximum  stresses 
being  7000  Ibs.  for  N  and  5000  Ibs.  for  T  (shearing  force  due 
to  middle  load  240). 

Ans.     d  =2.8+  in.  (about  3  ins.). 

12.  A  steel  bar  of  rectangular  section,  18  ft.  long,  1  in. 
thick,  and  8  ins.  deep,  is  under  tension  of  80,000  Ibs.  and  a 
bending  moment  of  13,230  Ib.-ins.     What  is  the  maximum 
stress  in  the  bar? 

Ans.     11,240  Ibs.  per  sq.  in. 

13.  A  wooden  beam,  8  ft.  long,  9  ins.  deep,  and  10  ins.  wide, 
is  under  compression  of  40,000  Ibs.  and  a  bending  moment  of 
48,000  Ib.-ins.     What  is  the  maximum  stress? 

Ans.     804J  Ibs.  per  sq.  in. 


CHAPTER   VI. 

SHEARING  STRESS  IN  BEAMS. 

31.  In  the  preceding  chapters  we  have  made  use  of  the 
simplest  kinds  of  loads.     In  tension  and  compression  we  can 
vary  the  intensity  of  the  stress  only  by  increasing  or  dimin- 
ishing the  load  as  it  has  already  been  applied;  in  torsion  also 
we  can  change  the  value  of  the  twisting  moment  only  by  vary- 
ing the  amounts  of  the  forces  forming  it  or  by  changing  the 
length  of  the  arm;  but  to  get  the  stresses  due  to  bending  and 
shearing  we  can  apply  loads  in  a  number  of  different  ways. 
We  have  also  been  neglecting  the  action  of  the  force  of 
gravity  on  our  rod,  which  is  of  importance,  as  in  the  case  of 
large  shafts  or  beams  whose  bearings  or  supports  must  be 
spaced  with  reference  to  their  weights;  besides  the  supporting 
forces'  for  a  beam  loaded  in  any  way  must  be  known  in  order 
to  find  the  shearing  and  bending  stresses. 

32.  Obviously,  as  we  have  assumed  in  Chapter  IV,  the 
supporting  forces  for  a  weightless,  horizontal  beam,  loaded 

W 

with  a  weight,  W,  exactly  at  its  middle,  are  each  — ,  or  half 

£t 

the  load;  if,  however,  we  take  into  consideration  the  weight 
of  the  beam  and  put  several  loads  upon  it  at  different  places, 
the  conditions  of  equilibrium  require  that  the  sum  of  the 
moments  of  all  the  forces  acting  on  the  beam  about  any  axis 
must  be  zero.  All  the  forces  acting  are  the  loads,  the  weight 
of  the  beam  and  the  supporting  forces,  all  acting  in  the  same 
plane.*  If  we  take  moments  about  an  axis  through  one  of 

*  If  the  forces  do  not  all  act  in  the  same  plane  those  acting  in  one 
plane  may  be  considered  separately,  and  the  resultant  of  the  several 
supporting  forces  in  the  different  planes  found.  The  supports,  however, 
must  be  strong  enough  in  the  directions  of  each  of  these  planes  to  sus- 
tain the  forces  acting  in  those  planes. 

52 


SHEARING  STRESS  IN  BEAMS.  53 

the  supports,  the  moment  of  that  supporting  force  will  be 
zero,  and  the  other  suporting  force,  being  now  the  only 
unknown  quantity,  is  readily  found  by  equating  to  zero  the 
algebraic  sum  of  the  moments  of  all  the  forces;  for  example, 
it  is  required  to  find  the  supporting  forces  of  a  beam  20  ft. 
long,  weighing  10  Ibs.  per  ft.,  and  supported  at  the  ends, 
carrying  150  Ibs.  6  ft.  from  the  left  end;  300  Ibs.  11  ft. 
from  the  left  end;  and  750  Ibs.  16  ft.  from  the  left  end  (see 
Fig.  24).  The  weight  of  the  beam  is  200  Ibs.  and  acts  at  its 

150  Ibs.          300  Ibs  750  Ibs 


Fig.  24. 

center  of  gravity,  which  in  this  case  is  at  the  middle  of  the 
beam.  If  we  take  moments  about  an  axis  through  the  left 
end  of  the  beam,  we  eliminate  the  supporting  force  P,  whose 
moment  about  this  axis  is  zero;  our  equation  for  equilibrium 
will  then  be 

200  X  10  +  150  X  6  +  300  X  11  +  750  X  16  -  Q  X  20  =  0, 

from  which  Q  =  910,  and  P  is  obviously  the  difference  be- 
tween the  sum  of  the  weights  and  Q,  or  P  =  490  Ibs^  In  the 
above  the  weights  act  at  the  points  indicated  and  are  known 
as  concentrated  loads.  If  the  loading  is  continuous  the  effect 
is  considered  as  acting  at  the  center  of  gravity,  as  it  has  been 
taken  for  the  weight  of  the  'beam  in  the  above. 

33.  Having  the  supporting  forces,  we  can  now  find  the 
shearing  stress  on  any  transverse  section  of  a  beam.  Refer- 
ring to  Fig.  25,  and  at  first  considering  the  beam  weightless, 
if  HH  is  any  section  between  the  supporting  force  P  and  the 


54 


STRENGTH  OF  MATERIAL. 


first  load  W^  there  would  clearly  be  a  tendency  for  the  two 
parts  of  the  beam  to  move  as  shown,  the  left  part  remaining 
stationary  and  the  right  part  being  pushed  down  by  the 
weights  Wl}  W2,  and  W3;  in  other  words  there  is  in  any 
loaded  beam  tangential  or  shearing  stress  along  any  trans- 
verse section.  The  magnitude  of  this  stress  (see  Chap.  II)  is 
equal  to  the  load  on  one  side  of  the  section  divided  by  the 
area  of  the  section.  In  the  beam  of  Fig.  25,  the  only  load  on 


Fig.  25. 

the  left  of  the  section  HH  is  the  supporting  force  P,  there- 
fore the  shearing  stress  on  any  section  between  P  and  the 

p 
first  load  TFX  is  —  •  *     Had  we  taken  our  section  between 

A 

Wl  and  W2  we  would  have  had  two  loads  on  the  left  of  the 
section  and  acting  in  opposite  directions,  so  that  the  shear- 

P  —  W 

ing  stress  would  have  been  — - — ,  and  so  on  across  the 

A. 

beam,  showing  a  drop  as  each  load  is  passed.  We  will  assume 
as  positive  the  conditions  as  in  Fig.  25,  the  tendency  being 
for  the  part  of  the  beam  to  the  right  of  the  section  to  move 
down;  then  let  us  construct  a  curve  of  shearing  force  using 
for  ordinates  the  values  as  found  above.  For  example,  Fig.  26 
is  a  beam  loaded  as  shown.  Taking  the  left  end  as  origin, 
the  total  shearing  force  at  that  end  is  equal  to  the  support- 
ing force  there,  or  S.  F.  =  P  =  29  Ibs.,  which  it  remains 

*  In  Chapter  II  we  took  our  shearing  forces  indefinitely  near  together 
to  get  pure  shear,  that  is,  shear  without  bending.  Shear  and  bending 
are  closely  connected,  so  that  it  is  difficult  to  produce  the  one  without 
the  other. 


SHEARING   STRESS   IN   BEAMS. 


55 


until,  as  we  move  the  section  to  the  right,  we  get  to  the  first 
load,  10  Ibs.,  which  acts  in  the  opposite  direction  to  the  sup- 
porting force;  so,  consequently,  as  we  pass  this  load  the  total 
shearing  stress  drops  to  19  Ibs.,  where  it  remains  until  we 


10  Ibs       20  Ibs 


30  Ibs 


Q=31I 


29        C        ? 


-X 


-31 
I 


Fig.  26. 

come  to  the  20-lb.  load.  Here  it  drops  to  -  1  Ib.  At  this- 
point  the  tendency  for  the  right  part  of  the  beam  to  move 
down  ceases,  as  the  sum  of  the  loads  on  the  left  side  of  a 
section  from  this  point  on  will  be  greater  than  the  supporting 
force  at  the  left  end.  Passing  the  30-lb.  load,  the  shearing 
stress  again  drops,  this  time  to  —  31  Ibs.,  and  the  curve  of 
shearing  force  will  be  the  series  of  steps  ABCDEFGH. 

These  values,  with  their  signs,  are  for  the  left  side  of  the 
section;  on  the  right  side  the  stresses  are  equal  to  those 
found  but  opposite  in  direction.  Had  we  used  the  right  end 
for  origin  and  the  left  side  down  as  the  positive  direction,  our 
curve  would  have  been  a  series  of  steps  down  to  the  left,  with 
numerical  values  the  same  as  before. 

34.  If  we  take  the  weight  of  the  beam  or,  what  is  the  same 
thing,  consider  the  beam  in  the  preceding  article  to  be  uni- 
formly loaded  all  along  its  length,  the  value  of  the  shearing 


56 


STRENGTH  OF  MATERIAL. 


force  will  change  at  each  point  as  we  move  our  section  to  the 
right.  Let  Fig.  27  represent  a  beam  so  loaded;  then  if  the 
ordinate  of  the  line  AB  represents  the  load  on  unit  length  of 
beam,  A B  will  be  the  load  curve,  being  drawn  below  the  beam 


Fig.  27. 


\% 

\ 

B 

w               W, 

W2                            3 

c 

m~    T"-:                     ^\         •-.    - 

Fig.  28. 

or  having  negative  ordinates  because  the  load  acts  down. 
The  supporting  forces  are  now  each  half  the  total  load,  and  at 
the  left  end  the  shearing  force  equals  the  supporting  force 
P;  but  as  we  move  the  section  to  the  right  we  reduce  the 
shearing  force  by  the  amount  of  the  load  between  the  section 


SHEARING  STRESS   IN  BEAMS.  57 

and  the  supporting  force,  the  S.  F.  being  equal  to  P  —  wx, 
where  w  is  the  load  per  unit  length,  and  x  the  distance  from 
the  left  end.  The  shearing  curve  will  therefore  drop  steadily, 
as  in  the  figure.  The  S.  F.  curve  is  CD.  If  we  have  in 
addition  to  the  uniform  load  several  concentrated  loads,  the 
curve  of  shear  will  be  as  in  Fig.  28,  with  a  drop  at  each  con- 
centrated load.  If  drawn  to  scale,  these  curves  will  give  the 
shearing  force  at  any  point  of  a  beam  by  measuring  the  ordi- 
nate  at  that  point.  As  a  GENERAL  RULE,  then,  the  shearing 
force  at  any  transverse  section  of  a  loaded  beam  is  equal  to 
the  algebraic  sum  of  all  the  loads  acting  on  one  side  of  that 
section. 

35.  Hereafter  we  will  use  the  left  end  of  the  beam  for  the 
origin  in  the  equations  for  curves  of  any  kind.  With  con- 
tinuous loads,  L,  the  load  at  any  point  will  vary  in  some 
way  with  the  distance  from  the  origin;  for  example,  a  beam 
"a"  ft.  long  carries  a  load  which  uniformly  increases  from 
zero  at  the  origin  to  w  Ibs.  per  ft.-run  at  the  right  end.  Here 
the  load  per  ft.-run  at  a  distance  x  from  the  origin  would  be 

—  >  found  from  the  ratio  —  =  — ,  and  the  equation  of  the 
a  x       a 

wx 

load  curve  (w  acting  downward)  would  be  L  = For  a 

a 

uniform  load  of  w  Ibs.  per  ft.-run,  L  =—  w.  The  equations 
for  load  curves  are  easily  found  and  are  very  useful,  for  let 
us  notice  the  relation  between  the  loads  and  the  shearing 
force  at  any  section.  As  a  general  formula  for  concentrated 
loads,  we  found  the  S.  F.  at  any  section  to  be  equal  to 
P  —  %W.  With  a  beam  loaded  with  any  continuous  load,  if 
L  be  the  load  per  unit  length,  then  Ldx  is  the  load  on  any 
elementary  length,  dx,  of  the  beam,  and  the  total  load  from 
the  origin  to  a  section  at  any  distance,  x,  from  the  origin  is 

(*x 

I   Ldx,  the  value  of  L  being  given  by  the  equation  to  the 
load  curve.     Now  the  value  of  this  integral  is  the  algebraic 


58  STRENGTH  OF  MATERIAL. 

sum  of  all  the  loads  to  the  left  of  the  section,  and  that  is 
also  our  definition  of  shearing  force.  As  a  general  rule,  then, 
for  any  continuous  load  the  shearing  force  is 

S.  F.  =   C *Ldx. 


In  using  the  formula  we  must  remember  that  the  constant  of 
integration  is  the  value  at  the  origin  of  the  quantity,  repre- 
sented by  the  integral  in  this  case,  that  is,  the  value  of  the 
supporting  force  P. 

For  concentrated  loads  we  must  still  use  S.  F.  =  P  —  2  W, 
and  if  a  continuously  loaded  beam  supports  also  a  number  of 
concentrated  loads  we  must  find  the  shearing  forces  sepa- 
rately, and  get  the  total  shearing  force  by  algebraically  adding 
the  results  (principle  of  superposition).  In  this  latter  case 
also  the  supporting  forces  for  both  the  continuous  and  con- 
centrated loads  must  be  found  and  used  separately. 

Examples: 

1.  A  beam  15  ft.  long  is  supported  at  the  ends  and  carries 
4  tons  5  ft.  from  the  left  end,  1  ton  8  ft.  from  the  left  end, 
and  J  ton  10  ft.  from  the  left  end.     Find  the  supporting 
forces  and  draw  the  shearing  curve. 

Ans.     P  =  3.3  tons.     Q  =  2.2  tons. 

2.  A  spar  20  ft.  long  is  supported  at  the  ends  and  loaded 
with  500  Ibs.,  4  ft.;  250  Ibs.,  9  ft.;  and  900  Ibs.,  18  ft.  from 
the  left  end.     Find  the  supporting  forces  and  draw  the  shear- 
ing curve. 

Ans.     P  =  627.5  Ibs.     Q  -  1022.5  Ibs. 

3.  A  plank  16  ft.  long  is  laid  across  a  ditch  and  a  man 
weighing  192  Ibs.  walks  across  it.     Find  the  shearing  curve 
when  he  is  3  and  when  he  is  8  ft.  from  the  left  end. 


SHEARING  STRESS   IN   BEAMS.  59 

4.  A  weight  of  384  Ibs.  is  placed  5  ft.  from  the  left  end  of 
the  above  plank.     What  is  the  shearing  force  at  a  point  6£  ft. 
from  the  left  end  when  the  man  is  8  ft.  from  there? 

Ans.      -  24  Ibs. 

5.  A  beam  18  ft.  long  has  a  uniform  load  of  50  Ibs.  per 
ft.-run.      Draw  curve  of  shearing  force  and  give  value  at 
points  7  ft.  and  13  ft.  from  the  left  end. 

Ans.     At  7  ft.  100  Ibs.;  at  13  ft.  -  200  Ibs. 

6.  On  the  beam  of  example  5,  weights  of  300  Ibs.  and  500 
Ibs.  are  placed  6  ft.  and  12  ft.  respectively  from  the  left  end, 
in  addition  to  its  uniform  load.     Draw  S.  F.  curve  and  give 
value  of  S.  F.  8  ft.  from  left  end. 

Ans.     116§  Ibs. 

7.  A  beam  5  ft.  long  has  its  left  end  fixed  in  a  wall  and 
supports  a  load  of  1000  Ibs.  at  its  free  end.     What  is  the 
S.  F.?  and  draw  curve. 

8.  The  beam  of  example  7  has  a  distributed  load  of  100 
Ibs.  per  ft.-run.     What  is  the  maximum  S.  F.?  and  draw  the 
curve. 

9.  An  oak  beam  15  ft.  long  and  1  ft.  square  floats  in  sea 
water.     It  is  loaded  at  the  center  with  a  weight  which  will 
just  immerse  it  wholly.     Draw  curve  of  S.  F.,  and  give  maxi- 
mum value.     (35  cu.  ft.  of  sea  water  weighs  1  ton;  1  cu.  ft. 
of  oak  weighs  48  Ibs.) 

Ans.     Maximum  S.  F.  =  120, Ibs. 

10.  A  pine  beam  20  ft.  long  and  1  ft.  square  floats  in  sea 
water,  and  is  loaded  at  the  middle  with  a  weight  which  will 
just  immerse  it.     Draw  a  curve  of  S.  F.  and  give  value  5  ft. 
from  left  end.     (Cu.  ft.  pine  weighs  39  Ibs.) 

Ans.     S.  F.  5  ft.  from  left  end  =  125  Ibs. 


60  STRENGTH  OF   MATERIAL. 

11.  A  beam  20  ft.  long,  supported  at  the  ends,  carries  a 
load  which  uniformly  increases  from  0  at  the  left  end  to  50 
Ibs.  per  ft.-run  at  the  right.     Draw  the  curve  of  S.  F.,  and 
find  its  maximum  value,  also  find  the  point  of  the  beam 
where  its  value  is  zero. 

Ans.     Maximum    value  =  333J    Ibs.     Value    zero    where 
x  =  11.5+  ft. 

12.  The  buoyancy  of  an  object  floating  in  the  water  is  0 
at  the  ends  and  increases  uniformly  to  the  center,  while  its 
weight  is  0  at  the  center  and  increases  uniformly  to  the  end. 
Draw  curve  of  S.  F.,  and  give  maximum  value. 

Ans.     Maximum  S.  F.  =  —  • 

4 


CHAPTER  VII. 

CURVES  OF  BENDING  MOMENTS  AND  SHEARING  FORCE. 

36.  In  Chapter  II  it  has  been  shown  that  when  a  beam  is 
loaded  there  are  horizontal  stresses  set  up  on  any  trans- 
verse section,  these  stresses  being  compressive  on  one  side 
and  tensile  on  the  other  side  of  the  neutral  plane.  Art.  14 
of  the  same  chapter  shows  that  the  moment  about  the  neu- 
tral axis  of  the  external  forces  on  one  side  of  any  section 
must  be  equal  to  the  moment  about  the  neutral  axis  of  the 
stresses  on  the  same  side  of  the  section.  Definition:  The 
bending  moment  at  any  section  is  the  moment  about  the 
neutral  axis  of  that  section  of  all  the  external  forces  on  one. 
side  of  that  section. 


By  definition  then  the  bending  moment  at  the  section  HH 
of  a  beam  loaded  as  in  Fig.  29  would  be 

M  =  Px  -  Wl  (x  -  x,)  -  W2(x  -  x2), 

or  finding  the  line  of  action  of  the  resultant  of  all  the  forces 
W^  W2,  etc.,  and  calling  its  distance  from  the  section  d 

M  =Px  -  2 

which  is  true  for  any  kind  of  load. 

61 


62  STRENGTH  OF   MATERIAL. 

If  we  now  consider  the  increment  AM  which  the  bending 
moment  receives  if  we  take  our  section  a  distance  Ax  to  the 
right,  the  bending  moment  about  the  neutral  axis  of  the 
new  section  will  be 

M  +  AM  =  P  (x  +  Ax)  -  STP  (d  +  Ax) 
=  Px  -  ZWd  +  Ax  (P  -  STT), 

but  Px  —  2Wd  is  the  original  bending  moment  equal  to  M, 
and  (P  —  2  FT)  is  our  formula  for  the  shearing  force  (F) 
at  any  section,  so 

M  +  AM  =  M  +  FAx,  or  AM  =  FAx, 

and  passing  to  the  limit  dM  =  F  dx;  integrating,  we  have 
for  a  general  formula  for  bending  moment 

M  =/F  dx. 

This  equation  is  true  for  any  kind  of  load,  but  care  must 
be  taken  to  use  the  correct  constant  of  integration,  which  is 
the  value  of  M  at  the  origin.  This  for  beams  free  at  the 
origin  is  zero,  but  if  a  beam  is  fixed  (prevented  from  moving 
in  any  way)  at  the  origin,  there  is  a  bending  moment  there. 
We  will  devote  the  rest  of  this  chapter  to  some  examples  of 
shearing  force  and  bending  moments  in  beams  loaded  and 
supported  ia  different  ways. 

37.  A  beam  supported  at  the.  ends  is  loaded  as  shown  in 
Fig.  30.  Find  the  curves  of  shearing  force  and  bending 
moment.  By  definition  the  S.  F.  =  P  —  SPF,  by  which  we 
get  the  curve  of  S.  F.  to  be  ABCDEF  (Fig.  30,  a).  Consid- 
ering each  load  separately,  the  supporting  forces  for  the 
25-lb.  load  are  P  =  17.5,  Q  =  7.5,  and  the  S.  F.  curve  is 
A  BCD  (Fig.  30,  6).  The  35-lb.  load  gives  supporting  forces 
p  =  y;  Q  =  28,  and  the  S.  F.  curve  is  ABCD  (Fig.  30,  c). 
If  we  add  algebraically  the  ordinates  given  by  these  curves 
at  any  point  distant  x  from  the  origin  we  will  get  the  ordi- 
nates for  the  S.  F.  curve  for  that  point  given  in  Fig.  30,  a. 


251bs 


351bs 


(63) 


64  STRENGTH  OF  MATERIAL. 

Obtaining  the  ordinates  by  definition,  the  curve  of  B.  M. 
for  the  beam  with  these  loads  is  given  by  OGHX  in  Fig.  30,  a. 

Let  us  get  separately  the  curves  of  B.  M.  due  to  each  load. 
For  the  25-lb.  load  we  get  by  definition  the  curve  in  Fig. 
30,  d,  the  maximum  B.  M.  being  directly  under  the  load  and 
the  curve  being  positive  at  all  points.  For  the  35-lb.  load 
we  get  the  curve  in  Fig.  '30,  e,  this  curve  also  being  positive 
at  all  points.  If  we  add  together  the  ordinates  given  by 
these  curves  at  any  point  distant  x  from  the  origin  we  will 
get  the  ordinate  of  the  curve  of  B.  M.  for  that  point  as  given 
in  Fig.  30,  a.  Now  by  definition,  the  bending  moment  at 
the  middle  of  the  beam,  for  example,  is 

24.5  X  5  -  25  X  2  =  72.5  Ib.-ft. 

and  by  the  formula  fF  dx  it  is  -  .5  X  5  =  -  2.5,  which 
is  evidently  incorrect,  but  is  due  to  the  fact  that  having 
taken  the  two  concentrated  loads  together,  the  shearing 
force  of  one  being  positive  and  the  other  negative,  we  have 
not  multiplied  the  total  shearing  force  by  x.  The  moments 
of  the  two  shearing  forces  cause  bending  in  the  same  direc- 
tion, therefore,  if  we  neglect  the  negative  sign  (which  we 
have  assumed  to  indicate  direction  only)  the  shearing  force 
at  this  point  will  be  7  +  7.5  =  14.5,  which  multiplied  by  5 
gives  72.5  as  before.  With  many  loads  to  get  the  shearing 
force  separately  would  be  a  tedious  operation,  so  it  is  better 
to  make  it  a  rule  to  get  the  S.  F.  and  B.  M.  for  concentrated 
loads  from  the  definition.  We  will  find  no  trouble  with 
beams  having  continuous  loads  unless  there  are  concentrated 
loads  in  addition,  in  which  case  we  get  the  curves  due  to  the 
continuous  loads  by  formula,  and  those  due  to  the  concen- 
trated loads  by  definition,  and  apply  the  principle  of  super- 
position. 

38.    A  beam  10  ft.  long  supported  at  the  ends  is  loaded 
with  a  uniform  load  of  25  Ibs.  per  ft.-run.     Find  the  curves 


CURVES  OF  BENDING   MOMENTS.  65 

of  S.  F.  and  B.  M.     In  this  case,  a  continuous  load,  our  for- 
mula is  very  convenient. 

L  =     -  w  =  -  25  Ibs. 

F  =  /Ldx  =  -  25  x  +  C. 
C  being  the  S.  F.  at  the  origin  is  equal  to  P  =  125  Ibs. 

.-.  F  =  125  -  25  x.  (1) 


Fig.  31. 

(1)  is  the  equation  to  the  curve  of  S.  F.  This  being  an 
equation  to  the  first  degree  is  a  straight  line  and  plots  as  the 
line  AB  of  Fig.  31.  The  bending  moment  is 


M  =  /Fdx  =/(125  -  25  x)  dx, 


or, 


M  -=  125  x 


25  x2 


[(7=0, 


(2) 


C  is  zero  because  B.  M.  is  zero  at  the  origin. 

(2)  is  the  equation  of  the  curve  of  B.  M.,  and  being  of  the 
second  degree  is  a  conic  (parabola)  and  plots  as  OCX  of 
Fig.  31.  The  maximum  B.  M.  is  at  the  middle  of  the  beam 
where  the  S.  F.  is  zero.  To  get  values  for  either  S.  F,  or 


66  STRENGTH  OF  MATERIAL. 

B.  M.  at  any  section  of  the  beam,  substitute  for  x  the  dis- 
tance of  the  section  from  the  origin  in  equation  (1)  or  (2) 
respectively. 

39.  A  beam  5  ft.  long,  fixed  at  the  left  end,  with  the 
right  end  unsupported,  carries  a  weight  of  50  Ibs.  on  the 
right  end.     Find  the  curves  of  S.  F.  and  B.  M. 

In  this  case  the  left  end  supports  the  whole  load,   so 
P  =  50,  and  the  shearing  force  by  definition  (concentrated 

loading)  is  F  =  50  Ibs.,  the 
curve  being  the  straight  line 
A  B  of  Fig.  32.  The  bend- 
ing moment  by  definition  is 
M  =  Px,  which  being  an  equa- 
tion of  the  first  degree  is  a 
straight  line,  the  ordinates 
varying  from  zero  to  250  Ib.-ft. 
Q2  This  would  give  us  a  line  in- 

clined upward  from  0,  but  we 
know  the  greatest  B.  M.  is  at  the  origin.  Now  notice  that 
the  curvature  of  this  beam  is  just  the  opposite  of  that  of  a 
beam  supported  at  the  ends,  the  center  of  curvature  being 
below  the  beam  in  this  case,  while  it  is  above  a  loaded  beam 
supported  at  the  ends.  In  fact  if  we  turn  Fig.  32  "upside 
down"  we  will  have  just  one-half  of  the  beam  supported  at 
the  ends  and  loaded  with  2  P  in  the  middle,  W  being  one  of 
the  supporting  forces;  so  this  kind  of  bending  is  called  neg- 
ative, and  instead  of  the  curve  of  B.  M.  inclining  upward 
from  0,  it  inclines  downward  from  X  as  in  the  figure.  The 
curve  plots  directly  if  we  take  the  origin  at  X  and  move  the 
section  to  the  left,  for  this  arrangement  is  the  same  as  a 
beam  twice  as  long  supported  in  the  middle  and  loaded 
with  50  Ibs.  at  both  ends. 

40.  A  beam  5  ft.  long  fixed  at  the  left  end,  with  the 
right  end  unsupported,  carries  a  uniform  load  of  25  Ibs.  per 
ft.-run.     Find  the  curves  of  S.  F.  and  B.  M, 


CURVES  OF  BENDING  MOMENTS. 


67 


A  continuous  load;   so  we  will  use  the  formula. 

L  =     -  w  =     -25  Ibs. 

F  =  /Ldx  =     -  25  x  +  C. 

C  being  the  S.  F.  at  the  origin  is  equal  to  P,  =125  Ibs. 

.-.  F  =  125  -  25  x.  (1) 


Fig.  33. 


(1)  being  an  equation  of  the  first  degree  is  a  straight  line 
and  plots  as  AX  in  Fig.  33.     The  bending  moment  is 


M  =  /Fdx  =/(125  -25  x)  dx  =  125  x  - 


25  x- 


Cr 


Here  Ct  is  not  zero.  But  we  know  there  is  no  bending  moment 
at  the  right  end  of  the  beam,  so  we  substitute  M  =  0  and 
z=5,  and  solve  for  C1; 


0  =  125  X  5  - 


+  Cv 


from  which  Cl  =  —  312.5;  substituting  this  value  of  Cl  we 
get 

^jti   *U 

(2) 


25  x2 
M  =  l25x 312.5 


for  the  equation  of  the  curve  of  B.  M.,  which  being  of  the 
second  degree  is  a  conic  (parabola)  and  plots  as  XC  in 
Fig.  33. 

This  equation  gives  a  maximum  negative  value  of  B.  M. 
at  the  origin,  where  we  know  it  should  be. 


68 


STRENGTH  OF   MATERIAL. 


41.  A  beam  10  ft.  long,  supported  at  the  ends,  is  loaded 
with  a  uniform  load  of  25  Ibs.  per  ft. -run  and  a  concentrated 
load  of  100  Ibs.  4  ft.  from  the  left  end.  Find  the  curves  of 
S.  F.  and  B.  M. 


Fig.  34. 

For  the  uniform  load  (Art.  38)  the  curve  of  S.  F.  is  AB 
(Fig.  34).  For  B.  M.  the  curve  is  01 X. 

For  the  concentrated  load  the  S.  F.  curve  is  CDEF,  and 
the  curve  of  B.  M.  is  OJX. 

Adding  algebraically  the  respective  ordinates,  we  get  the 
S.  F.  curve  for  both  loads  to  be  GHMN,  and  for  B.  M.  the 
curve  of  both  loads  is  OLX. 

42.  A  beam  10  ft.  long  is  supported  at  the  ends  and  has  a 
load  uniformly  increasing  from  zero  at  the  left  end  to  100 
Ibs.  per  ft.-run  at  the  right  end.  Find  curves  of  S.  F.  and 
B.  M.  Here  (see  Art.  35) 


L  =  - 


100  x 
10 


-  10  x, 


10 


S.  F.  =/Ldx  =/-lOxdx  = —  +  C, 


CURVES  OF  BENDING   MOMENTS. 


69 


where  C  being  the  S.  F.  at  the  origin  is  equal  to  P.  We 
can  find  P  by  the  method  used  in  the  example  at  the  end  of 
Art.  32,  as  follows:  In  Fig.  35  OL  is  the  load  curve.  The 


Fig.  35. 

center  of  gravity  of  the  load  then  acts  at  §  the  length  of  the 

beam  from  0,  and  the  total  load  is  100  X  -     =  500  Ibs.; 

fit 

therefore,  taking  moments  about  0, 

500  X  §  .  10  -  Q  X  10  =  0; 

and  Q  =  333J,  the  other  supporting  force  being  500  -  333 J; 
or  P  =  166§.     But  we  can  get  P  in  another  way,  for 


S.  F.  =  P  -  5  x2 


and 


B.M.=   (Fdx=  (  (P  -  5x2)dx=Px 
J  J 


5x5 


Cl  is  equal  to  zero,  for  the  bending  moment  at  the  origin  is 
zero,  and  M  is  also  zero  at  the  right  end  of  the  beam,  so  that 
if  we  substitute  x  =  10  and  solve  the  equation 

5  X  1000 
P  X  10  --  -  -  =  0 


70  STRENGTH  OF  MATERIAL. 

we  get  P  =  166§  as  before.     Putting  this  value  of  P  in  the 
equation  for  S.  F.  and  B.  M.  we  get 

S.  F.  =  166§  -  5  x2  (1) 

and 

B.  M.  =  166§  x  -  —  ,  (2) 

3 

which  equations  give  the  curves  ACB  for  shearing  (Fig.  35) 
and  ODX  for  bending. 

If  we  put  (1)  equal  to  zero  and  solve  for  x  we  get  the  point 
on  the  beam  where  the  shearing  force  is  zero;  and  if  we  put 
this  value  of  x  in  (2)  we  will  get  the  maximum  B.  M.,  for 
the  maximum  B.  M.  occurs  where  the  S.  F.  is  zero  (see  Art. 
28),  for  by  the  principles  of  maxima  and  minima,  the  first 
derivatives  of  the  B.  M.  with  respect  to  x,  put  equal  to  zero, 
will  give  us  the  shearing  equation  (1)  equal  to  zero, 


d  (BM)  I  500        10 

_b '  =  166§  -  5z2=0;   from  which  x=\/-  —  = , 

dx  V  3  x  5      V3 

the  value  of  x  where  B.  M.  is  a  maximum. 


Examples  : 

1.  A  beam  10  ft.  long,  supported  at  the  ends,  carries  a  load 
of  1000  Ibs.  4  ft.  from  the  left  end.     Find  curves  of  S.  F. 
and  B.  M. 

2.  A  beam  5  ft.  long,  fixed  at  the  left  end  and  unsupported 
at  the  right  end,  carries  1000  Ibs.  at  the  right  end.     Find 
curves  of  S.  F.  and  B.  M. 

3.  A  beam  10  ft.  long,  supported  at  the  ends,  carries  a  uni- 
form load  of  100  Ibs.  per  ft.-run.     Find  curves  of  S.  F.  and 
B.  M.;  give  values  at  ends  and  center. 

Ans.     S.  F.,  ends  ±  500;  center  0.     B.  M.,  ends  0;  center 
1250  Ib.-ft. 


CURVES   OF  BENDING   MOMENTS.  71 

4.  A  beam  5  ft.  long,  fixed  at  the  left  end,  unsupported  at 
the  right  end,  carries  a  uniform  load  of  100  Ibs.  per  ft.-run. 
Find  curves  of  S.  F.  and  B.  M.  and  give  values  at  ends. 

Ans.     S.  F.,  left  end  500  Ibs.;  right  end  0.     B.  M.,  left  end 
-  1250  lb.-ft.;  right  end  0. 

5.  A  beam  10  ft.  long,  supported  at  the  ends,  carries  400 
Ibs.  4  ft.  from  the  left  end,  and  600  Ibs.  6  ft.  from  the  left 
end.     Find  curves  of  S.  F.  and  B.  M.  and  give  values  at  the 
ends  and  center. 

Ans.     S.  F.,  left  end  480  Ibs.;  right  end  520;  center  80. 
B.  M.,  left  end  0  Ibs.;  right  end  0;  center  2000  lb.-ft. 

6.  What  is  the  longest  steel  bar  of  cross-section  of  1  sq.  in. 
that  can  be  supported  at  its  center  without  being  perma- 
nently  bent,   the  greatest   allowable   bending  moment  for 
the  bar  being  2000  Ib.-ins.? 

Ans.     19      ft. 


7.  A  beam  20  ft.  long,  supported  at  the  ends,  carries  2000 
Ibs.  5  ft.  from  the  left  end,  and  5000  Ibs.  4  ft.  from  the  right 
end.     Find  curves  of  S.  F.  and  B.  M.  and  give  values  at  the 
ends  and  at  the  loads. 

Ans.  S.  F.,  left  end  2500  Ibs.;  right  end  4500  Ibs.;  be- 
tween loads  500  Ibs.  B.  M.,  end  0;  first  load  12,500  lb.-ft.; 
second  load  18,000  lb.-ft. 

8.  If  the  beam  of  example  7  were  loaded  with  200  Ibs.  per 
ft.-run,  find  the  curves  of  S.  F.  and  B.  M.  and  give  values  at 
ends  and  center. 

Ans.  S.  F.,  ends  ±  2000  Ibs.;  center  0.  B.  M.,  ends  0; 
center  10,000  lb.-ft. 

9.  A  round  steel  rod  of  2  ins.  diameter  can  only  withstand 
a  bending  moment  of  6  ton-ins.     What  is  the  greatest  length 
of  such  a  rod  which  will  just  carry  its  own  weight  when  sup- 
ported at  the  ends? 

Ans.     29.2  ft. 


72  STRENGTH  OF  MATERIAL. 

10.  A  beam  20  ft.  long,  supported  at  the  ends,  carries  a 
uniformly  distributed  load  of  5  tons,  and  a  concentrated  load 
of  5  tons,  4  ft.  from  the  left  end.     Find  curves  of  S.  F.  and 
B.  M.  and  give  values  at  the  center.     Where  is  the  greatest 
bending  moment?  and  give  its  value. 

Ans.     S.  F.,   center  -  1   ton.     B.  M.,   center  22^  ton-ft. 
B.  M.,  maximum  24^  ton-ft.,  6  ft.  from  left  end. 

11.  A  pine  beam  20  ft.  long  and  1  ft.  square  floats  in  sea 
water.     It  is  loaded  at  the  center  with  a  weight  just  sufficient 
to  immerse  it  wholly.      Find  curves  of  S.  F.  and  B.  M.  and 
give  maximum  values.     A  cu.  ft.  of  pine  weighs  39  Ibs.,  of 
sea  water  64  Ibs. 

Ans.     Maximum  S.  F.,  250  Ibs.     Maximum  B.  M.,   1250 
Ib.-ft. 

12.  A  beam  54  ft.  long,  supported  at  the  ends,  is  loaded 
with  15  cwt.  per  ft.-run,  for  a  distance  of  36  ft.  from  the  left 
end.     Find  the  curves  of  S.  F.  and  B.  M.     What  is  the  maxi- 
mum B.  M.  and  the  B.  M.  at  6,  12,  and  36  ft.  from  the  left 
end? 

Ans.     B.  M.  maximum  =  216  ton-ft.;  B.  M.6  =  94.5  ton- 
ft.;  B.  M.12  =  162  ton-ft.;  B.  M.36  =  162  ton-ft. 

13.  A  steel  beam  5  ft.  long  is  fixed  at  one  end,  unsup- 
ported at  the  other  end,  and  is  of  rectangular  section  2  ins. 
wide  and  3  ins.  deep.     What  weight  at  the  free  end  will  de- 
stroy the  beam  if  the  limiting  stress  is  24  tons  per  sq.  in.? 

Ans.     1.2  tons. 

14.  The  buoyancy  of  a  floating  object  is  0  at  the  ends,  and 
increases  uniformly  to  the  center,  while  the  weight  is  0  at 
the  center  and  increases  uniformly  to  the  ends.     Find  the 
curves  of  S.  F.  and  B.  M.  and  give  maximum  values  in  terms 
of  the  displacement,  Z>,  and  the  length,  /,  of  the  object. 

D                                        ID 
Ans.     S.  F.  maximum  = ;  B.  M.  maximum  = —  - 


CHAPTER   VIII. 

SLOPE  AND  DEFLECTION. 

43.  The  slope  at  any  section  of  a  loaded  beam  is  the  angle 
between  the  tangent  to  the  neutral  line  at  that  section  and  the 
straight  line  with  which  the  neutral  line  would  coincide  if  the 
beam  were  not  bent;  or,  slope  is  the  angle  between  our  axis 
of  X  and  the  tangent  at  any  section  to  the  curve  into  which 
the  beam  is  bent. 

The  deflection  at  any  section  of  a  loaded  beam  is  the  dis- 
tance from  the  axis  of  X  to  the  point  where  the  neutral  line 
pierces  that  section;  or,  it  is  the  ordinate  at  that  section  of 
the  neutral  line  (see  Fig.  36). 

From  Chapter  IV  we  have  the  general  formula  for  bending, 

M      E  ^  El 

—  =  • —  ,  from  which  R  =  —  > 

E  being  a  constant  and  /  also,  in  this  case,  as  we  are  consid- 
ering beams  of  uniform  cross-section;  M,  of  course,  varies  at 
different  sections  of  the  beam. 

By  calculus  the  formula  for  the  radius  of  curvature  at  any 
point  of  a  curve  given  by  its  rectangular  equation  is 


dx2 
Equating  these  two  values  of  R  we  have 


TM  n 

M  d?y 

dx* 
73 


74  STRENGTH  OF  MATERIAL. 

which  by  integration  will  give  us  the  equation  to  the  curve 
into  which  the  beam  is  bent.  As  this  integration  would  be 
somewhat  complicated,  and  as  in  properly  built  structures  the 
dimensions  of  the  different  pieces  of  material  are  such  that 
the  bending  is  very  slight,  we  can  without  appreciable  error 

dy 

simplify  the  operation  considerably  as  follows:  —  being  the 

ax 

tangent  of  the  angle  that  the  bent  beam  at  any  point  makes 
with  the  axis  of  X,  is  a  very  small  fraction,  and  being 
squared  in  equation  (1)  becomes  so  small  that  it  may  be  neg- 
lected, in  comparison  with  unity,  so  that  equation  (1)  for  all 
practical  purposes  becomes 

El         I  „.  d*y 

—  ----  —  ;  or,  El  —  •  =  M; 

M       d2y  dxz 

dx2 

and  integrating  this  equation  we  get 

EI—=/Mdx,  (2) 

dx 

which  will  give  us  the  tangent  of  the  slope  at  any  point  when 
we  substitute  for  M  its  value  in  terms  of  x  as  found  in  Chap- 
ter VII,  and  integrate.  The  integral  f  Mdx  is  called  the 
slope  function,  and  we  will  designate  it  by  S.  When  S  is 
divided  by  El  we  have  the  tangent  of  the  angle  of  slope. 
If  we  write  equation  (2) 


and  integrate  again  we  get 

Ely  =/Sdx.  (3) 

This  is  the  equation  of  the  curve  into  which  the  beam  is  bent, 
and  the  value  of  y  given  by  this  equation  is  the  ordinate 
of  the  neutral  line  at  any  section  distant  x  from  the  origin. 
y  is  usually  negative. 


SLOPE  AND  DEFLECTION.  75 

44.  In  integrating  to  get  values  for  equations  (2)  and  (3) 
we  must  not  forget  the  constants  of  integration. 

If  a  beam  is  fixed  at  the  origin,  the  slope  there  is  zero,  but 
a  beam  supported  at  the  ends  and  loaded  will  bend  into  a 
curve  like  that  of  Fig.  36.  Clearly  the  slope  is  greatest  at 
the  ends  and  there  will  be  a  constant  of  integration  for  equa- 
tion (2).  We  will  usually  know  from  the  way  in  which  the 
beam  is  loaded  a  value  of  x  at  which  the  slope  is  zero. 


o 

H 

•#-  X  •*» 

^                                    -^ 

X 

-4 

' 

'    ~  "-^l^LL.            ~       ~^^~-^~^ 

f 

Q 

Fig.  36. 


For  example,  with  a  uniform  load  the  beam  of  Fig.  36 
would  bend  so  that  at  its  middle  the  slope  would  be  zero; 
therefore,  if  we  substitute  (having  integrated  the  right  mem- 
ber) for  x,  in  equation  (2),  half  the  length  of  the  beam  and 
put  the  equation  equal  to  zero,  we  can  solve  for  C,  the  con- 
stant. If  we  do  not  know  a  value  of  x  where  the  slope  is 
zero,  we  will  know  a  point  where  the  deflection  is  zero  (one 
of  the  supports  for  example),  and  substituting  in  equa- 
tion (3)  the  value  of  x  for  this  point,  solve  for  the  constant 
of  integration  of  equation  (2).  An  example  will  be  solved 
illustrating  the  method. 

There  is  usually  no  difficulty  in  finding  the  constant  of 
integration  for  equation  (3),  because  the  deflection  of  a  beam 
at  the  origin,  whether  the  beam  be  fixed  or  supported  there 
and  no  matter  how  loaded,  is  zero.  Of  course  a  beam  could 
be  propped  up  somewhere  along  its  length,  so  that  the  left 
end  would  not  rest  on  its  support. 


76  STRENGTH   OF   MATERIAL. 

45.  A  beam  10  ft.  long,  supported  at  the  ends,  is  loaded 
with  a  uniform  load  of  25  Ibs.  per  ft.  -run.  Find  the  slope 
and  deflection. 

Here  the  bending  is  perfectly  symmetrical.  Let  Fig.  36 
represent  the  bent  beam,  0  being  the  origin  and  OX  the 
position  of  the  neutral  line  before  the  load  was  applied,  the 
dotted  line  representing  its  position  after  the  application  of 
the  load. 

p  =  Q  =  125. 

L  =  -  25, 

F  =    -  25x  +  C  =     -25.T  +  125.     [C  =  P], 

oe  ~2 

M  =  -    -   +  125  x+  [C  =  0, 


dx  6  2 

Knowing  the  bending  is   symmetrical,  -1-  =  0,  where  x  =  5 

dx 

(the  middle),  so 


In  case  we  do  not  know  the  bending  is  symmetrical,  we 
carry  Cl  down  through  the  next  integration  for  deflection, 
which  gives 

Elv  =  ~  2^f  +  nr  +  CtX  +  [C'  =  °-      (2) 

C2  is  zero,  for  the  deflection  is  zero  at  the  origin.  The  deflec- 
tion is  also  zero  where  x  =  10  (the  other  support),  and 
substituting  this  value  we  get 

25  X  10000       125  X  1000 

o  =  ----  r~  +  -  -  -  +ioc,, 

from  which  Cl  =  -  1041  §  as  before,  and  equation  (1) 
becomes 

«*?!        +  !?.  -10411  (3) 


SLOPE  AND   DEFLECTION.  77 

from  which  by  substituting  the  abscissa  of  any  section  of 
the  beam  for  x  and  dividing  the  result  by  El  we  get  the 
tangent  of  the  angle  of  slope  for  that  section.  Substitut- 
ing the  value  of  Cl  in  equation  (2)  gives 


(4) 


from  which  the  ordinate  of  the  neutral  line  (curve  of  bend- 
ing) at  any  section  may  be  found.  In  finding  the  value  of 
this  ordinate,  or  of  the  slope,  care  must  be  taken  to  use  the 
same  linits  throughout;  for  example,  if  we  have  a  steel 
beam  and  E  is  given  as  30,000,000,  it  is  in  pounds  per 
square  inch;  we  must  therefore  use  x  in  inches,  find  /  for 
the  section  in  inch  units,  and  reduce  the  bending  moment 
to  inch-pounds;  this  will  give  us  the  deflection  in  inches. 

It  is  convenient  to  solve  this  problem  by  using  letters  to 
represent  the  numerical  data  and  to  make  the  substitutions 
later  in  the  equation  which  gives  the  desired  result;  for 
example,  in  the  above  problem  let  the  load  per  foot-run 
equal  "w"  and  the  length  of  the  beam  "a,"  then 

L  =  —  w, 

wa 

F  =  -wx  +  C  =  -wx  +  P  =  -wx-l  --  , 

2i 

wx2      wax 
M  =  -  —  +—+[Cl=0, 

dy          wx3      wax2 


/a\3 
(  —  ) 
\2J 


a 

wa .  ^  . 

wa{ 


wx       wax       wax 


78 


STRENGTH  OF  MATERIAL. 


By  this  latter  method,  the  particular  result  desired  may 
be  obtained  without  doing  the  numerical  work  required  for 
the  other  equations.  In  the  above  beam  the  maximum 

7/Y7 

slope  is  at  the  ends,  and  the  tangent  of  the  angle  is  --  —  • 

The  maximum  deflection  occurs  at  the  middle  of  the  beam, 

5  iva4 
and  there  j,  =     - 


46.  A  beam,  supported  at  the  ends,  carries  a  single  con-.. 
centrated  load  W  at  a  distance  a  from  the  left  end  and  b 
from  the  right  end.  Find  equation  for  slope  and  deflection. 


a+b 


a+b 


Fig.  37. 


In  this  problem  the  beam  will  be  bent  as  shown  in  Fig.  37, 
and  the  elastic  curve  will  consist  of  two  branches,  the  part 
from  0  to  the  weight  and  that  from  the  weight  to  the  other 
end  of  the  beam.  The  shearing  force  on  each  part  will  be 
constant,  but  the  values  will  be  different  and  they  will  have 
different  signs.  WTe  must  consider  the  two  parts  sepa- 
rately. With  the  origin  at  0,  the  bending  moment  for  any 
section  HH  to  the  left  of  the  load  W  is 

Wbx 
M  =  -     -, 

a  +  b 

Wbx2 


dy  _ 

dx      2  (a  +  6) 


Ely 


Wbxs 


6  (a  +  6) 


[C2  =  0. 


SLOPE  AND  DEFLECTION.  79 

For  the  part  to  the  right  of  the  load  W  we  will  take  our 
origin  at  the  right  end,  then,  letting  ;r,  be  the  distance  from 
the  right  end  to  any  section  H'H' ',  the  bending  moment  will 
be 

M-       Qx 

~X    ~ 


the  sign  being  negative  because  the  moment  tends  to  turn 
the  right  end  of  the  beam  counter  clockwise,  or  in  the  oppo- 
site direction  to  that  of  the  other  end;  we  have  then  for  the 
right  end 

*  — *&, 

a  +  b 
d,i  Wax* 


dx  2  (a  +  b) 


If  we  move  the  origin  for  these  equations  back  to  the  left 
end  of  the  beam,  by  substituting  {x  —  (a  +  b)}  for  xl  we 
get 

Wa  (a  +  6  -  x) 


M  = 


a  +  b 


dy  Wa 


dx  2  (a  +  b} 


Now  for  the  section  under  W  the  slope  and  deflection  are  the 
same,  using  either  the  equations  for  the  part  of  the  curve  to 
the  left  of  W,  or  those  for  the  part  to  the  right  of  W,  so  we 


80 


STRENGTH  OF   MATERIAL. 


can  put  the  corresponding  values  equal  to  each  other  when 
x  =  a 


Wba- 


2  (a  +  6) 


Wba3 
6  (a  +  6) 


Wab2 


2  (a  +  6) 


(1) 


6  (a  +  6) 
or,  C>  +  C/6  = 


(6  -  a) 


(2) 


From  the  simultaneous  equations  (1)  and  (2)  we  get 

Wab  (a  +  2  6)  ,  _  Tf  afc  (2  a  +  6) 

1-          6  (a  +  6)    /  8         '  "       6  (a  +  6) 

and  substituting  these  values  of  C,  and  (?/  we  ge* 

for  the  right  end, 
£;/<%==_TFa(a+6-a;)3 


for  the  left  end, 
Wbx*         Wab  (a  +  2  6) 


2  (a  +  6)  6  (a  +  6) 


EIy  = 


6  (a  +  6) 


6  (a  +  6) 


dx 


Ely 


2  (a  +  6) 


6  (a  +  6) 
Wa  (a  +  b  -  xY 
6  (a  +  6) 

TFa6(2a  +  b)(a  +  b  -  x] 
6  (a  +  6) 


and  from  either  set  of  these  equations  we  get  the  deflection 
at  the  load  to  be 


3  El  (a  +  6) 

Since  the  load  is  not  at  the  middle  of  the  beam,  the  maxi- 
mum deflection  will  occur  in  the  longer  segment,  and  at  the 
section  of  maximum  deflection  the  slope  will  be  zero;  there- 


SLOPE  AND  DEFLECTION.  81 

fore,  supposing  a  to  be  greater  than  b  if  we  put  the 
equation  of  the  slope  for  the  part  of  the  curve  to  the  left  of 
W  equal  to  zero  and  solve  for  x  we  will  get  the  abscissa  of 
the  section  of  maximum  deflection;  and  substituting  this 
value  of  x  in  the  equation  for  the  deflection  will  give  us  the 
maximum  deflection,  as  follows: 


Wbx2          Wab  (a  +  2  b)  a  (a  +  2  6) 


and  this  value  of  x  substituted  in 

Wbx3          Wab  (a  +  2  6)  x 

Ely  =  --- 
6  (a  +  b)  6  (a  +  b) 

shows  the  maximum  deflection  to  be 


Wb          /a  (a  +  2 
ymax~  ~  ' 


3  El  (a  +  6) 
or,  if  a  =  6, 

Wa3 


/a  (a  +  2  b)\ 
(          3         / 


Umax 


6  El 


Examples: 

1.  A  steel  beam,  10  ft.  long,  supports  a  concentrated  load 
of  25  tons  at  its  middle.     What  is  the  deflection  under  the 
load  if  /  in  inch-units  is  84.9  and  E=  30,000,000?     Find 
equation  of  elastic  curve. 

Ans.     .077  —  ins. 

2.  A  steel  beam  of  /  section,  20  ft.  long,  8  ins.  deep, 
3  ins.  wide,  with  flanges  and  web  each  J  in.  thick,  is  used  to 
support  a  floor  weighing  200  Ibs.  per  sq.  ft.     The  beams  are 
supported  at  the  ends  and  spaced  3  ft.  apart.     What  is  the 
maximum  deflection?     E  =  30,000,000.     Find  equation  of 
elastic  curve. 

Ans.     1.3  —  ins. 


82  STRENGTH  OF  MATERIAL. 

3.  A  beam,  supported  at  the  ends,  carries  a  uniform  load 
of  150  Ibs.  per  ft. -run.     It  is   10  ft.  long,  6  ins.  deep,  and 
4    ins.    wide.     E  =  1,200,000.     Find    equation    of    elastic 
curve  and  the  maximum  deflection. 

Ans.     .39  —  ins. 

4.  A  beam,  12  ft.  long,  8  in.  deep,  and  12  in.  wide,  is  sup- 
ported at  the  ends  and  carries  a  load  of  1200  Ibs.,  3  ft.  from 
the  left  end.     E  =  1,200,000.     Find  the  elastic  curve  and 
the  maximum  deflection. 

Ans.     .102  +  ins. 

5.  A  steel  beam,  60  ft.  long  and  weighing  100  Ibs.  per  ft., 
carries  a  concentrated  load  of  14,400  Ibs.  at  the  middle.     If 
/  in  inch-units  is  1160,  find  the  maximum  deflection.     E  = 
30,000,000. 

Ans.     4.055  —  ins. 

6.  Beams,  20  ft.  long,  support  a  floor  weighing  100  Ibs. 
per  sq.  ft.     The  beams  are  spaced  4  ft.  apart.     Supposing 
the  section  of  the  beam  to  be  square,  find  the  side  of  the 
square  in  order  that  the  deflection  may  not  exceed  \  in. 
E  =  700  in.-tons. 

Ans.     Side  is  12.18  ins. 


CHAPTER  IX. 

SLOPE  AND  DEFLECTION  —  Continued. 

47,  Example:  A  beam  of  length  a  is  fixed  at  the  left 
end  and  unsupported  at  the  right.  It  is  loaded  with  a  uni- 
form load  of  w  Ibs.  per  ft.-run.  Find  the  equation  of  the 
elastic  curve. 

L  =  —  w, 

F  =  —  wx  +  C  =  —  wx  +  wa  (P  =  whole  load), 

M  =      —  +  wax  +  [Cl  =      —  (Cl  is  found  in  Art.  40), 
dii          wx3      wax2      wa'2x 

Eldx=--&+^r  -2-  +  [c>  =  ° 

(beam  fixed  at  origin), 
wx 4      wax3      wa2x2 

E/J/=~2?  +  -6"       ^+^=0 

(beam  fixed  at  origin). 

The  maximum  deflection  obviously  occurs  where  x  =  a. 

wa4 


I/max   = 


8  El 


Let  us  suppose  the  free  end  of  the  above  beam  is  propped 
up  to  the  same  level  as  the  fixed  end.  The  prop  now  sup- 
ports some  of  the  load,  so  we  do  not  know  P. 

L  =  —  w, 

F  =  -  wx  +  C  =  -  wx  +  P, 

M  =  -  ^f  +  Px  +  Ct. 

a 

83 


84 


STRENGTH   OF  MATERIAL. 


We  know  the  bending  moment  at  the  end  supported  by  the 
prop  is  zero,  so  putting  M  =  0  and  substituting  x  =  a, 


from  which 


, 
>--- Pa. 


Substituting  this  value  of  Cl  we  get 


M  = 

wx* 

f  Px  H 

wa* 
h           -  Pa, 

2 

2 

m*a 

wx3 

D,y,2 

w;a2a: 

Pax  - 

Jdx 

6 

2 

2 

wx* 

Px3 

wa2£2 

i 

Pax2 

,      .          .  .   . 
(beam  fixed  at  origin), 


(beam  fixed  at  origin). 

We  can   now  find  P,  for  the  deflection  is  zero  where  x  =  a, 
and  all  the  other  constants  are  determined. 


wa* 


Pa: 


load 


Fig.  38. 


from  which 


P  = 


5  wa 


and  substituting  this  value  in  the 
equations  will  give  us  the  curves 
desired  and  the  slope  and  deflection. 
The  curves  are  shown  plotted  in  Fig. 
38,  AB  being  the  curve  of  S.  F.,  and 


CDX  that  of  bending  moment.     The  B.  M.  at  the  origin  is 


wa 


negative  and  equal  to .     The  greatest  positive  B.  M. 


SLOPE  AND   DEFLECTION.  85 

occurs  where  the  shearing  force  is  zero  at  x  =  — ,  and  substi- 
tuting this  value  in  the  equation  for  B.  M.  gives  the  greatest 

positive  B.  M.  equal  to  —    —  .    The  maximum  B.  M.  is  there- 
128 

fore  the  negative  one,  and  the  beam  will  break,  if  overloaded, 
at  the  fixed  end. 

It  will  be  noticed  that  the  B.  M.  is  zero  at  the  free  end, 
and  also  at  a  point  between  there  and  the  fixed  end.  Putting 
the  equation  for  B.  M.  equal  to  zero  and  solving  for  x  gives 

a  a 

x  =  a  and  x  =  - .     The  point  where  x  =  -  is  called  a  virtual 
4  4 

joint,  because,  if  a  beam  loaded  and  supported  in  this  way 
had  a  hinge  at  this  point  the  bending  moment  would  not 

cause  it  to  turn.     Putting  —  =  0  and  solving  for  x  gives 

CttC 

x  =  .58  a  and  a  value  greater  than  the  length  -of  the  beam. 
Where  x  =  .58  a  the  slope  is  zero,  and  this  is  the  point  of 
maximum  deflection.  Substituting  this  value  of  x  in  the 
equation  for  deflection  gives 

.006  wa4 


Compare  the  curve  of  B.  M.  for  this  beam  with  that  in 
Fig.  33.  As  soon  as  the  prop  under  the  free  end  of  this  beam 
begins  to  bear  a  part  of  the  load  there  exists  a  positive  B.  M., 
and  the  virtual  joint,  which  is  then  near  the  right  end  of  the 
beam,  moves  toward  the  fixed  end,  as  the  right  end  is  propped 
up  and  the  load  on  the  prop  increases;  but  the  results  due  to 
change  in  conditions  are  readily  followed. 

48.    A  beam  a  ft.  long  is  fixed  at  both  ends  and  carries 

W 
a  concentrated  load  W  at  its  middle.      Here  P=  Q=  — 


86 


STRENGTH  OF  MATERIAL. 


W 


.*.   S.  F.  =  —  and  is  constant  from  the  ends  to  the  load,  but 

has  opposite  directions  on  opposite  sides  of  the  load;  there- 
fore, as  in  Art.  44,  we  must  consider  the  two  parts  of  the 
beam  separately.  To  the  left  of  the  load,  then, 


M 


(C  is  not  known), 


0 


(beam  fixed  at  end). 


dx         4 

The  loading  being  symmetrical,  the  slope  at  the  middle  is 

a 

zero ;  substituting  x  =  —  i 
2 


The  equation  of  the  curve  of  bending  to  the  left  of  the  load, 
then'is  Wx      Wa 


/\ 

B\ 

/ 

y.H  ; 

K\  

p            / 

\     Q' 

\ 

"fe 

Fig.  39. 

This  is  an  equation  of  the  first  degree,  and  therefore  repre- 
sents a  straight  line  (EF  in  the  figure).  We  know  the  B.  M. 
is  symmetrical,  and  that  the  curve  for  the  right  end  will  have 
equal  ordinates,  but  of  opposite  sign  (the  curve  being  FG  in 


SLOPE  AND  DEFLECTION.  87 

the  figure).  Now  it  is  clear  we  cannot  embrace  both  of  these 
straight  lines  in  one  equation  of  the  first  degree,  and  for  this 
reason  we  must  consider  the  two  parts  separately.  Before 
proceeding,  we  will  notice  that  the  bending  moment  has  the 
same  value  with  different  signs  at  the  ends  and  in  the  middle. 
This  shows  that  the  beam  is  as  likely  to  break  at  the  ends  as 
in  the  middle  if  overloaded,  and  also  that  the  virtual  joints 
are  at  quarter  span. 

Wx      Wa 


The  maximum  deflection  is  at  the  middle  and  equal  to 

Wa3 
'/max~  IQ2YI' 

49.  The  Cantilever  Bridge.  —  In  the  cantilever  bridge  the 
joints  are  placed  at  the  points  where  the  bending  moment 
would  be  zero  if  the  bridge  were  continuous  over  the  whole 
span.  We  will  consider  a  beam  fixed  at  both  ends,  having 
two  fixed  joints  and  carrying  a  uniform  load  of  w  Ibs.  per 
ft.-run.  There  being  joints  at  F  and  G  (Fig.  40)  we  must 
consider  the  part  FG  as  a  separate  beam  supported  at  the 
ends,  the  load  on  it  being  equally  divided  and  supported  at 
F  and  G  by  the  beams  OF  and  GX.  From  the  conditions 
we  could  assume  P  and  Q  with  this  loading  to  be  equal  to 

w  (a  +  6  +  c) 

—  -  —     —  j  but  we  can  get  their  values  in  another  way: 

L  =  —  w, 

F  =  -  wx  +  C  =  -  wx  +  P, 


88  STRENGTH  OF  MATERIAL. 

The  bending  moment  at  the  joint  is  zero,  so 


0= 


w  (a  +  6)2 


and  from  equations  (1)  and  (2)  we  get 


+C19 


=  -(2a  +  6),andC1--  -- 

^ 


substituting, 


w 


=  -  wx  +  -(2  a  +  6), 

,4 


(2) 


The  joints  being  where  B.  M.  is  zero,  the  curves  of  S.  F.  and 
B.  M.  will  be  continuous  as  shown  in  Fig.  40,  but  for  slope 
and  deflection  we  must  consider  each  part  as  a  separate  beam, 


Fig.  40. 

the  parts  at  ends  having  in  addition  to  their  uniform  load  a 
concentrated  load  at  their  ends,  F  and  G,  equal  to  half  the 
total  load  on  the  middle  part.  For  equal  strength,  the  bend- 


SLOPE  AND   DEFLECTION. 


89 


ing  moments  at  the  ends  and  middle  of  the  above  beam  should 
have  equal  values,  and  this  requires  a  to  be  equal  to  c  and 


wa  wb2 

(0  +  6)  =  — 


whence 


=  ?r;  or,  4  a 


=  2  62;  or,  (2  a 


or  the  square  of  the  whole  span  must  equal  2  62,  and  the  joint 

span 

must  be  distant  from  the  middle  of  the  beam  ——=.  - 

2V2 

.    50.   Traveling  Loads.  —  If  a  load  W  moves  from  left  to 
right  across  a  beam  of  length  a  the  positive  shearing  force 


=     («-*> 


Fig.  41. 


for  any  position  of  W  is  equal  to  P  (Fig.  41).  P  is  ob- 
viously greatest  when  W  is  just  over  it,  and  as  W  moves  across 
the  beam  P's  value  drops  uniformly  until  when  W  is  just 
over  Q,  P's  value  is  zero.  We  can  then  represent  the  change 
of  the  positive  S.  F.  by  the  line  AX,  and  by  the  same  reason- 
ing the  change  in  the  negative  S.  F.  would  be  represented 
by  OB.  Considering  the  B.  M.  we  know  that  for  any  posi- 
tion of  W  the  B.  M.  is  greatest  for  the  section  under  W  and 

W 
for  that  section  is  M  =  Px  —  —  (a  —  x)  x.    By  the  principles 


of  maxima  and  minima  we  can  find  the  value  of  x  for  which 


90  .  STRENGTH  OF  MATERIAL. 

M  is  a  maximum  by  putting  the  first  derivative  of  M  with 
respect  to  x  equal  to  zero  and  solving  for  x,  as 


dM      W 

—  =  —  (a  -  2  z)  =  0, 

dx        a 


which  gives 


the  value  of  x,  or  the  position  of  W  which  gives  the  maximum 
B.  M.  If  the  traveling  load  is  continuous,  such  as  a  train 
of  cars  crossing  a  bridge,  the  maximum  value  occurs  when, 
if  the  train  is  long  enough,  it  extends  completely  over  the 
bridge;  if  it  is  not  long  enough  for  this  the  maximum  positive 
S.  F.  occurs  when  the  whole  train  has  just  got  on  to  the  bridge 
and  the  maximum  B.  M.  when  the  middle  of  the  train  is  just 
over  the  middle  of  the  bridge. 

51.  Oblique  Loading.  —  Loading    is    said    to    be    oblique 
when  a  principal  axis  of  any  cross-section  of  a  beam  does 
not  lie  in  the  plane    of   the   external 
bending  moment. 

Let  Fig.  42  represent  the  section  of 
a  beam  and  let  the  arrow-head, 
marked  M,  represent  the  direction  of 
the  B.  M.  The  axis  of  X  is  perpen- 
dicular to  the  plane  of  the  paper 
through  0,  and  the  axes  of  Y  and  Z 
Fi  42  are  shown.  Resolve  M  into  compon- 

ents parallel  to  the  axes  of  Y  and  Z. 

The  direction  of  the  fiber  stress  due  to  bending  is  perpen- 
dicular to  the  plane  of  the  paper,  and  that  part  of  it,  due  to 
M  sin  a,  which  acts  normal  to  the  side  AD  of  the  section  is 

M  sin  a  :  z 

p.--  --  -> 


SLOPE  AND   DEFLECTION.  91 

which  is  obtained  by  substituting  M  sin  a,  the  component 
moment,  in  -  =  — ,  the  general  equation  for  bending,  Iy 

y     l 

being  the  moment  of  inertia  of  the  section  about  the  axis  of 
Y,  and  z  the  distance  of  the  line  AD  from  the  axis  of  Y. 
The  normal  fiber  stress  perpendicular  to  the  plane  of  the 
paper  on  AB  is  M  cog  a 


Applying  the  principle  of  superposition  the  total  fiber  stress 
at  A  would  be 

M  sin  a  .  z      M  cos  a  .  y 

p  =  PV+  PZ=     —j—  -j- 

52.  The  work  done  in  bending  a  beam  is  evidently  equal 
to  M6,  where  M  is  a  uniform  bending  moment  and  6  is  the 
angle  of  slope  at  the  ends.  If  I  is  the  length  of  the  beam  and 

R  the  radius  of  the  curve  into  which  it  is  bent,  then  6  = , 

2  R 

ME  El 

and  from  —  =  —  we  have  R  =  —-  •     Hence  the  work  done 
I      R  M 

MH 

by  a  uniform  bending  moment  is  equal  to •    But  bend- 

2  El 

ing  moments  are  seldom  uniform,  so  for  a  variable  bending 
moment  i 

Work=  — /Af'«fa. 


Examples: 

1.  A  dam  is  supported  by  a  row  of  uprights  fixed  at  their 
base  and  their  upper  ends  held  vertical  by  struts  sloping 
at  45°.  Water  pressure  varies  as  the  depth,  or  L  =  —  wx. 
Find  the  equation  for  deflection.  Length  of  upright  a. 
What  is  the  thrust  on  the  struts? 

Ans,     Thrust  =  f  horizontal  pressure  of  .water, 


92  STRENGTH  OF   MATERIAL. 

2.  A  railroad  is  inclined  at  30°'  to  the  horizontal.     The 
stringers  are  10.5  ft.  apart  and  the  rails  are  1  ft.  inside  the 
stringers.     The  ties  are  8  in.  deep  and  6  in.  wide.     The  load 
transmitted  by  each  rail  to  one  tie  is  10  tons.     What  is  the 
maximum  normal  stress  in  each  tie? 

Ans.     6428.8  Ib.-ins. 

3.  A  beam,  2. a  ft.  long,  fixed  at  the  ends,  is  uniformly 
loaded  with  w  Ibs.  per  ft.-run.     Find  the  maximum  deflec- 
tion and  the  virtual  joints. 

1/Jfl  ^ 

Ans.     Maximum   deflection  =  -  -      Virtual    joints 

/-  24  El 

where  x  =  a  (1  ±  v  J). 

4.  A  beam,  a  ft.  long  and  fixed  at  the  ends,  carries  a 
uniformly  increasing  load  from  zero  at  the  left  end  to  w  Ibs. 
per  ft.-run  at  the  right.     Find  the  maximum  deflection  and 
the  virtual  joints. 

A    -if  1/7  ^ 

Ans.     Maximum  deflection  =  —  —  •     Virtual  joints 

a          2  a  3125  El 

where  x  =  -  and  — . 

5.  A   beam,   a   ft.  long,  fixed  at   one   end   and  the  other 
end   propped   up   to   the   same   level,    carries   a   uniformly 
decreasing  load  from  w  Ibs.  per  ft.-run  at  the  fixed  end  to 
zero  at  the  propped  end.     Find  the  equation  for  deflection, 
the  position  of  the  maximum  positive  B.  M.,  and  the  position 
of  the  virtual  joints. 

Ans.     Maximum   positive   B.  M.   where   x  =  a  \/$.     Vir- 
tual joints  where  x  —  a  >/|. 

6.  A  beam,  a  ft.  long,  fixed  at  both  ends,  carries  a  single 
concentrated  load  W  at  a  distance  d  from  the  left  end.   Find 
the  deflection  under  the  load. 

Wd3  (a  -  d)3 

Ans.     3  = 

3  Ela* 


SLOPE  AND   DEFLECTION.  93 

7.  If  the  load  of  the  beam  in  example  6  had  been  at  the 
middle  of  the  beam,  what  would  have  been  the  maximum 
deflection? 

Wa3 


Ans.      d  = 


192  El 


8.  A  single  load  of  50  tons  crosses  a  bridge  of  100-ft. 
span.     Draw  the  curves  of  maximum  S.  F.  and  B.  M.,  and 
give  the  values  of  those  quantities  at  half  and  quarter  span. 

9.  A  train,  weighing  1  ton  per  ft.  -run  and  112  ft.  long, 
crosses  a  bridge  of  100-ft.  span.     Draw  curves  of  maximum 
S.  F.  and  B.  M.,  and  give  values  at  half  and  quarter  span. 

10.  A  steel  shaft  carries  a  load  equal  to  k  times  its  own 
weight,  first  uniformly  distributed,  second  concentrated  at 
its  middle;   considering  it  as  a  beam  fixed  at  the  ends,  find 
the  distance  apart  of  the  bearings  that  the  ratio  of  deflec- 
tion to  span  may  be 


Ans.     (1)    Span  in  feet=  10.5 


3/ 
y 

* 


/C    ~T~    1 
3    /~J2 

(2)    Span  =  8.3  \/  —  (d  in  inches). 

»    K   +   T? 


CHAPTER   X. 
CONTINUOUS  BEAMS. 

53.  A  continuous  beam  is  one  which  extends  over  several 
supports.  In  this  chapter  we  will  consider  as  heretofore 
only  such  beams  as  have  a  uniform  cross-section  and,  as  in 
all  practical  constructions  the  supports  are  adjusted  so  as  to 
allow  as  little  strain  as  possible  in  the  material,  we  will 
assume  all  the  supports  to  be  at  the  same  level.  It  is  obvi- 
ous that  the  bending  moments  at  the  intermediate  supports 
will  be  negative,  as  at  these  points  the  conditions  are  similar 
to  those  of  a  beam  balanced  over  a  single  support.  If  the 
end  spans  are  short  we  may  have  the  supporting  forces  at 
the  ends  equal  to  zero,  and  if  the  ends  are  "  anchored  " 
(fastened  down  to  the  support)  we  may  have  a  negative  sup- 
porting force,  that  is,  one  acting  in  the  same  direction  as  the 


Fig.  43. 

loads.  In  Fig.  43  we  see  approximately  the  curves  of 
bending  moments  and  shearing  force  for  a  continuous  beam 
carrying  a  uniform  load  and  having  five  equally  spaced  sup- 
ports. There  are  two  virtual  joints  between  supports  ex- 
cept for  the  end  spans,  and  we  will  find  that  the  value  of  the 
deflection  in  each  span  is  somewhere  between  that  for  a 

94 


CONTINUOUS   BEAMS.  95 

uniformly  loaded  beam  of  span  length  supported  at  the  ends 
and  one  uniformly  loaded  of  span  length  fixed  at  the  ends. 
The  difficulty  in  working  with  continuous  beams  lies  in  find- 
ing the  supporting  forces,  for  the  usual  method  of  taking 
moments  will  not  do,  there  being  two  unknown  quantities 
in  each  equation,  and  the  equations  reducing  to  identities 
when  we  attempt  to  solve  them.  For  beams  having  only 
three  supporting  forces,  the  solutions  are  not  difficult,  as  will 
be  shown;  but  as  the  number  of  supports  increases  the  cal- 
culations become  more  and  more  tedious. 

54.  A  beam  of  length  2a  carries  a  uniform  load  and  is 
supported  by  three  equidistant  supports.  A  beam  of 
length  2a  carrying  a  uniform  load  of  w  per  unit  length  and 
supported  at  the  ends,  has  a  maximum  deflection  equal  to 

---  (see   Art.  45).     A  like  beam   supported    at   the 

ends  and  carrying  a  concentrated  load  R  at  its  middle  has 

R  (2a)3 

a  maximum  deflection  equal  to  -    -  (Art.  46).     Now 

48  El 

the  above  continuous  beam  may  be  considered  as  a  uni- 
formly loaded  beam  which  has  a  prop  at  its  middle  by  which 
the  middle  point  is  propped  up  to  the  same  level  as  the  end 
supports.  Obviously  then  the  deflection  at  any  point  will 
be  that  due  to  the  uniform  load  minus  that  due  to  the 
thrust  of  the  prop,  and  as  the  prop  deflects  upward  we  have 


~~  ~48W    ~  384  El 


or  R,  the  thrust  of  the  prop,  or  the  middle  supporting  force 


The  supporting  forces  at  the  ends  are  obviously  equal,  so 

w  (2a)  -  1  wa 
P  =  Q  =-  —  =  ft  wa. 


96  STRENGTH   OF  MATERIAL. 

Having  the  supporting  forces  we  have  now  no  difficulty  in 
getting  the  equations  for  slope  and  deflection,  for,  taking  the 
origin  at  the  middle,  the  B.  M.  for  any  section  distant  x  from 

L_I 


Fig.  44. 

the  origin  is,  by  definition  (remembering  that  the  part  of  R 

R\ 
which  is  due  to  the  loading  on  one  side  of  the  origin  is  — J, 


dy       Rx2       WX3  f  from  symmetry 

El—  =-  -  +  MQ  X  +  [C  =  0  .  .  .  .  <  the  slope  is  zero 

dx         46  r  at  the  origin. 

77,  T         "&         WX  X  \  deflection  is  zero 

J=~~  ~       +  M°         [   l=  ^ the 


Substituting  the  value  of  R  we  have  for  the  B.  M.  at  any 
section 

M  =  f  wax  -  -—  +  MQ. 

And  as  the  B.  M.  is  zero  at  the  ends,  we  have,  substituting 

wa2 
x  =  a,  the  value  of  M0  equal  to  -   — —  ;  or 

o 

wx2      wa2 
M  =  ft  wax  -  —  ; 


CONTINUOUS   BEAMS.  97 

putting  this  value  of  M  equal  to  zero  and  solving  for  x 

a 

gives  us  x  =  ±  —  as  the  position  of  the  virtual  joints.  In- 
tegrating we  get  the  equations  for  slope  and  deflection, 

dy         5  wx3       wa2x 

El  f-  =  —  wax2-  —  -  +  [C  -  0, 

dx        16  68 

and 

5          3      wx4      wa2x2 

48™        "  "24""       16         ^  l~ 

To  show  how  carefully  the  level  of  the  supports  must  be 
adjusted,  suppose  there  is  left  a  deflection  over  the  middle 
support,  equal,  let  us  say,  to  £  of  the  total  deflection  which 
would  occur  were  there  no  middle  support;  then 

1     5w  (2' a)4  _5w  (2  a)4      R  (2  a)3 
5       384  El          384  El          48  El  ' 

w  (2  a) 
from  which  R  = »  or  one-half  the  total  load.     If  this 

deflection  were  in  the  other  direction,  or  —  £,  we  would  have 
R  =  4  the  total  load.  Remembering  that  the  deflection  in 
any  case  is  very  small  when  we  see  that  a  variation. of  it  £ 
up  or  down  causes  R  to  change  from  j  to  J  the  total  load, 
we  can  understand  how  carefully  the  supports  must  be  ad- 
justed to  the  same  level. 

55.  With  concentrated  loads  we  must  proceed  in  a  differ- 
ent way.  Suppose  a  beam  which  is  supported  at  the  middle 
and  ends  carries  a  concentrated  load  W  midway  between  the 
supports, .  as  represented  in  Fig.  45.  Taking  the  origin  at 
the  middle  as  before  and  considering  the  right  half  of  the 
beam,  the  bending  moment  at  any  section  HH  between  the 
origin  and  the  load  is  (Art.  46) 

Q  (a  -  x).  (I) 


98  STRENGTH  OF   MATERIAL. 

Integrating, 

dy  Wax 

and 


For  a  section  between  W  and  the  end  of  the  beam 

dx2 
Integrating, 

-  ^T  +  (C  =  ?.  (5a) 


W 


^3 
-i 

Q=X«w 


Fig.  45. 


The  slope  under  the  load  is  the  same  in  both  branches  of  the 
curve;  therefore,  substituting  x  =  —  -  in  equations  (2)  and 
(5a)  and  equating  them  we  have 

Wa2      Wa2      Qa2      Qa2  _Qa2     Qa2  Wa2 

—£~      ~^~      ~2~~    ~g~=    ~2~     ~8~+        °        =        ~8~' 

Substituting, 

dy  Qx2       Wa2 


CONTINUOUS   BEAMS.  99 

Integrating, 

E-MCl  =  ,  (6a) 


The  deflection  under  the  load  is  the  same  in  both  branches, 
so  as  before,  using  equations  (3)  and  (6a)  we  have 

Wa^      TFo3       Qa3      Qa3  =  Qa3      Qa3      Wa3      Q 

'~16         48"          8      '  48   "      8    "   48  "     16          lf 
or, 

Wa3 
Ll=      48    ' 

Substituting, 

Qax2      Qx3      Wa2x      Wa3 


Now  the  deflection  at  the  right  support  is  equal  to  zero,  and 
putting  equation  (6)  equal  to  zero,  substituting  x  =  a,  solv- 
ing for  Q,  gives  Q  =  T6F  W.  From  symmetry,  P  equals  Q, 
and  R  must  support  the  rest  of  the  load,  hence 

R=2W-(P  +  Q)  =  V  W. 

The  methods  of  this  and  the  preceding  article  will  cover 
most  of  the  cases  of  continuous  beams  found  in  actual  prac- 
tice, but  there  are  many  cases  where  beams  have  more  than 
three  supports,  and  for  these  a  method  known  as  the 
"  Method  of  Three  Moments  "  must  be  employed.  This 
method  came  into  use  about  1857,  and  is  a  general  method 
for  obtaining  the  supporting  forces. 

56,  Theorem  of  Three  Moments.  —  Let  Fig.  46  represent 
the  part  of  a  beam  over  any  three  consecutive  supports, 
and  let  the  loading  be  uniform  between  adjacent  sup- 
ports. Take  the  origin  at  P  and  let  the  bending  moments 


100  STRENGTH  OF  MATERIAL. 

at  P,  R  and  Q  be  Mly  M2  and  M3  respectively,  the  distance 
between  supports  to  be  a  and  alt  the  loading  between  P  and 
R  be  w  Ibs.  per  unit  length,  and  between  R  and  Q  be  wl  Ibs. 


J 


L T— -«--- — 4—       — «— 

|p  ^  [R  |Q 

Fig.  46. 

per  unit  length;    then  the  bending  moment  at  any  section 
HH  between  P  and  R  is  (Art.  45) 

wxz 
M  =         — -  +  Px  +  C. 

But  C  in  this  case  is  equal  to  Mlf  so 

Integrating, 

El—  =  M.x  +  P  —  -  —  +  [C  =  ?,  (2) 

dx  26 

and 

B/y-ar^+P^-^+Cx  +  tc.-o 

6       24 

(level  supports).    (3) 

Now  2/  =  0  when  x  =  a,  hence 

ATa2       Pa3      im4  M.a     Pa2      i^a3 

0  =  — h +  Ca.    .'.  C  = +  — 

2  6          24  2          6         24 


CONTINUOUS  '-SSAMS.  101 


Substituting, 

dy                   Px2     wx3      M^a       Pa2  wa3 

dx~  lX      ~2~        6          2           6  24 

and 

M/Y»2  D/y*3               'jpy1^                 /I//"    /Tf  O*  P/Tf^'Y*  1/1/7     "7* 

.  ,*/  X^  •//              UUJU              ±  rJ.  4  Lt  Ju  JL   C6  Ju  Cc/C*   *(/ 


From  (1),  if  we  let  x  =  a,  we  get  M2  in  terms  of  Mlt  or, 
M2  =  M,  +  Pa  -  ^f  •  (6) 

jfc 

Now  if  we  imagine  the  beam  reversed  and  use  Q  for  the 
origin,  we  can  in  the  same  way  find  the  equation  for  bend- 
ing moment,  slope  and  deflection;  for  the  part  of  the  beam 
between  Q  and  R  and  over  R  the  values  of  the  bending 
moment,  slope  and  deflection  will  be  the  same  for  both 
branches  of  the  curve.  We  can  therefore  equate  these 
values  if  we  substitute  x  =  a  in  those  of  the  left  branch  and 
x  =  at  in  those  of  the  right.  Making  these  substitutions  and 
equating  we  get 

for  the  left  branch,  for  the  right  branch, 

Ml  +  Pa-^  =  M2;=M3  +  Qa1-^,       (7) 

M.a      Pa2     wa3 


,          .. 
=s,opeover/J    —     i-L  +  -Li..  (8) 

If  we  substitute  the  values  of  Q  and  P  from  equation  (7) 
in  equation  (8)  and  reduce  we  get 

7/V7  ^     I     7/5   //    ^ 

M,a  +  2  M2  (a  +  a,)  +  M3a,  =  -          "^  1  1  •  (9) 

This  is  known  as  the  equation  of  the  three  moments,  and 
by  using  different  sets  of  three  consecutive  supports  we  can 
get  as  many  equations,  less  two,  as  there  are  supports. 


102  STRENGTH  -Q&  MATERIAL. 

From  our  knowledge  of  cortditkms  at  the  ends  of  the  continu- 
ous beam  we  can  get  two  more  equations  involving  the 
moments  over  the  supports,  and  thus  having  as  many  equa- 
tions as  there  are  unknown  quantities  we  may  find  all  the 
supporting  forces. 

For  example,  let  the  beam  of  Fig.  43  be  supported  at 
equal  intervals  and  uniformly  loaded  with  w  Ibs.  per  unit 
length.  From  equation  (9) 


or, 


(1) 

(2) 
(3) 


The   bending   moment   is   zero   at   the   ends,   hence   M, 
M5=0.     From  (1)  and  (3) 


Substituting,  we  get  from  (2)  and  (1) 


From  (6),  using  the  first  three  supports, 
wa2  3  wo? 


Now  Q  is  equal  to  the  change  in  S.  F.  at  the  second  support. 
The  negative  part  which  is  due  to  the  load  to  the  left  of  the 
support  is  P  —  wa,  and  the  positive  part  is  given  by  the  equa- 


CONTINUOUS  BEAMS.  103 


tion  M3  =  M2  +  QR  X  a  -      —,  and  the  sum  of  the  two  parts 

ft 

gives 

Q  =  3  §  wa, 

In  the  same  way  R  =  f  |  wa,  or  from  symmetry  we  know 
S  =  Q  and  T  =  P,  so  the  total  load  less  (P  +  Q  +  T  +  S) 
=  R,  from  which  R  =  f f  wa. 


Examples : 

1.  A  continuous  beam  of  five  equal  spans  is  uniformly 
loaded  with  w  Ibs.  per  ft.-run.     If  the  beam  is  38  ft.  long, 
what  are  the  supporting  forces  and  the  bending  moments  at 
the  supports? 

43  37 

Ans.     P  =  3  w.     Q= — w.     R  =  —  w. 
5  5 

2.  Find  the  side  of  a  steel  beam  of  square  section  to  span 
four  openings  of  8  ft.  each,  the  total  load  per  span  being 
44,000  Ibs.  and  the  greatest  horizontal  fiber  stress  not  to 
exceed  15,000  Ibs.  per  sq.  in. 

Ans.     Side  =  5.66  in. 

3.  Find  the  depth  of  a  steel  beam  of  rectangular  section 
twice  as  deep  as  broad,  to  span  three  openings  each  12  ft. 
wide,  the  total  load  on  each  span  being  6000  Ibs.  and  the 
greatest  fiber  stress  allowed  being  12,000  Ibs.  per  sq.  in. 

Ans.     4.4  in. 

4.  A  continuous  beam  of  three  spans  is  loaded  only  on 
the  middle  span  with  a  uniform  load.     What  are  the  sup- 
porting forces? 

wl 

Ans.     At  ends 

20 


104  STRENGTH  OF  MATERIAL. 

5.  A  square  steel  beam,  /  =  268.9  in  inch-units,  is  36  ft. 
long  and  spans  four  openings,  the  end  spans  being  each  8  ft. 
and  the  middle  ones  each  10  ft.  Find  the  maximum  B.  M. 
of  the  beam.  What  will  be  the  uniform  load  per  foot  to 
make  a  maximum  fiber  stress  of  15,000  Ibs.  per  sq.  in.? 
261  w 


Ans. 


31 


6.  A  continuous  beam  carries  a  uniform  load.     If  the 
end  spans  are  each  80  ft.  what  will  be  the  length  of  the 
middle  span  in  order  that  the  B.  M.   at  its  middle   may 
equal  zero? 

Ans.     1=  67. 14  ft. 

7.  How  much  work  is  done  on  a  beam  10  ft.  long,  10  ins. 
deep  and  8  ins.  wide,  which  carries  a  uniform  load  of  250  Ibs. 
per  ft.-run? 

5 
Ans.     Work  = 


CHAPTER   XL 

COLUMNS  AND  STRUTS. 

57.  When  a  prismatic  piece  of  material,  several  times 
longer  than  its  greatest  breadth,  is  under  compression,  it  is 
called  a  column  or  strut;  a  column  being  such  a  piece  placed 
vertically  and  carrying  a  static  load,  and  all  others  being 
struts. 

A  column  or  .strut  of  material  which  is  not  homogeneous, 
or  one  on  which  the  load  does  not  act  exactly  inthe  geometric 
axis,  will  bend  or  buckle.  For  mathematical 
investigation  we  must  assume  our  material 
homogeneous,  and  instead  of  assuming  a  slight 
deviation  of  the  line  of  action  of  our  load  from 
the  geometric  axis,  we  will  assume  that  the 
column  has  first  been  bent  by  a  horizontal 
force,  then  such  a  load  applied  to  its  end  as 
will  just  keep  it  in  the  bent  form  after  the 
removal  of  the  horizontal  force.  The  assump- 
tions we  make  will  be  three:  (1)  the  column 
perfectly  straight  originally;  (2)  material  per- 
fectly homogeneous;  and  (3)  load  applied  ex- 
actly over  the  center  of  the  ends.  These 
conditions  are  never  exactly  fulfilled  in  practice, 


Fig.  47. 

so  a  rather  large  factor  of  safety  must  be    applied. 

58.  Eider's  Formula  for  Long  Columns.  —  We  will  choose 
a  column  with  rounded  or  pivoted  ends,  so  that  they 
will  be  free  to  move  slightly  as  the  column  is  bent.  Let 
Fig.  47  represent  such  a  column,  held  in  the  bent  position  by 
the  load  W.  Let  the  origin  be  at  the  center  of  the  column, 
the  axis  of  X  vertical  and  that  of  Y  horizontal ;  d  is  the  maxi- 

105 


106  STRENGTH   OF   MATERIAL. 

mum  deflection,  and  ?/  the  ordinate  of  the  neutral  line  at  any 
section  HH  distant  x  from  the  origin.  From  Art.  43  we  have 
the  general  equation  of  bending, 


The  bending  moment  for  the  section  HH  is,  from  the  figure, 

M  ==  W(d  -  y), 
so 

d^=Ti(8-y)- 

Multiplying  both  members  of  this  equation  by  2  —  and  inte- 
grating, we  get 


now  y  is  zero  at  the  origin  and   —  is  also  zero  there,  the 

tangent  to  the  curve  of  bending  being  parallel  to  the  axis  of 
X  at  that  point,  therefore  the  constant  of  integration,  being 
this  tangent,  is  zero. 

Extracting  the  square  root  of  both  members  of  equation 
(1)  and  transposing,  we  get 


_Jw 


-i-r;   a 

which  integrated  gives 


(2) 


C2  is  zero,  for  where  x  =  0,  y  =  0  and  cos~l  1=0. 


COLUMNS   AND  STRUTS. 


107 


Letting  I  equal  the  length  of  the  column,  when  x  =  -  ,  we 
have  y  =  d;  substituting  these  values  in  equation  (2)  we  get 


The  angle  whose  cosine  is  zero  is  - 


from  which  we  get 


—. 

El'%' 


'El 

"F 


which  is  Euler's  formula  for  long  columns  with  round  ends. 
Transposing  equation  (2)  we  get  for  the  equation  of  the 
elastic  curve 


y  =  d  ]  1  —  cos, 


59.  A  column  loaded  with  W  as  per  the 
above  formula  will  just  retain  any  deflection 
which  may  be  given  it,  therefore  this  value  of 
W  is  called  the  critical  load,  for  the  column  will 
straighten  out  if  there  be  any  decrease  in  this 
load,  and  for  any  increase  it  will  keep  on  bend- 
ing until  it  breaks.  This  is  the  load  then 
which  puts  the  column  in  neutral  equilibrium. 
The  bending  of  columns  is  perfectly  uniform,  so 
we  can  derive  from  the  formula  of  the  preced- 
ing article,  which  is  for  a  column  with  rounded 
or  pivoted  ends,  the  formula  for  columns  with 
one  or  both  ends  fixed.  Let  Fig.  48  represent 
the  elastic  curve  of  a  column  with  both  ends  fixed.  There 
will  be  two  points  of  inflection,  B  and  D,  and  as  the  bending 


108  STRENGTH  OF  MATERIAL. 

is  perfectly  uniform  these  points  will  be  at  quarter  the  length 
of  the  column  from  the  ends.  The  part  BCD  will  represent 
the  elastic  curve  of  a  column  such  as  we  considered  in  the 
preceding  article,  so  the  critical  load  of  a  column  with  fixed 
ends  will  be  the  same  as  for  a  column  of  half  its  length  with 
pivoted  ends,  or, 


The  part  BCDE  of  Fig.  48  would  approximately  represent 
the  elastic  curve  for  a  column  with  one  end  (E)  fixed  ;  so  the 
critical  load  for  it  would  be  the  same  as  for  a  column  of  two- 
thirds  its  length  with  pivoted  ends,  or 


This  latter  formula  is  not  quite  so  accurate  as  the  others,  for 
the  ends  are  not  in  the  same  vertical  line. 

It  has  been  shown  by  experiment  that  when  the  length  of  a 
pillar  is  greater  than  100  diameters  the  theoretical  values  are 
closely  approached,  while  with  shorter  lengths  these  values 
are  much  too  large. 

Columns  having  flat  ends,  if  the  ends  are  prevented  from 
lateral  movement,  are  considered  as  having  fixed  ends. 

60.  As  Euler's  formula  is  applicable  only  in  the  case  of 
very  long  columns,  another  formula  has  been  obtained  in 
different  ways  by  several  different  writers,  which  gives  more 
accurate  results  for  short  columns.  Obviously  a  column  may 
be  so  short  as  to  fail  by  crushing  alone,  in  which  case  the 
crushing  load  would  be  W  =  fA,  where  /  is  crushing  strength 
of  the  material  and  A  the  area  of  the  cross-section.  For 
ordinary  columns,  then,  W  must  lie  between  fA  and  the  value 
given  by  Eider's  formula. 

The  following  formula  is  most  used  for  short  columns. 
If  pl  be  the  compressive  stress  due  to  W,  and  p2  the  fiber 


COLUMNS  AND  STRUTS.  109 

stress  due  to  bending,  the  maximum  stress  will  be  Pi  +  p2', 
this  must  not  exceed  the  strength  of  the  material,  so  for  the 
limiting  stress,  /=  p1  +  p2. 

W 

If  A  is  the  cross-sectional  area  of  the  column,  pl  =  -—. 

A. 

p      M  My 

The  equation  of  bending,  —  =  —  ,  gives  us  p2  =  — -  ,  remem- 

<y 

bering  that  y  in  this  formula  is  the  distance  from  the  neutral 
plane  to  the  extreme  fiber  of  the  section, 

_W      My 
Z      ~T 

From  Fig.  48,  if  the  maximum  deflection  be  d,  we  have 
M—  Wd,  and  as  in  symmetrical  bending  d  varies  as  --  we 

y 

have  d  =  —  (the  constant  is  always  a  fraction,  so  may  be  put 

in  this  way) ;  substituting  these  values  and  remembering  /  is 
equal  to  A  multiplied  by  a  constant  squared,  or,  /  =  Ak2, 
we  have 

_  W      Wl2y_W  i         72 

A         cyl       A 
from  which 

Af 


W 


This  is  known  as  Rankin's  formula  for  columns  and  struts 
with  fixed  ends. 

61.   We  see  that  when  I  approaches  zero,  in  the  above 

formula,  W  approaches  Af,  and  as  -  increases,  W  will  ap- 

k 

proach  the  value  given  by  Euler's  formula. 


110  STRENGTH  OF   MATERIAL. 

The  formula  of  Art.  60  is  for  a  column  or  strut  with  fixed 
ends;  if  both  ends  are  rounded  or  pivoted,  we  must  divide  the 
value  of  c  by  4;  if  one  end  is  rounded,  c  is  divided  by  2,  so 
that  the  formulae  for  different  methods  of  end  supports  are 

Af 
Flat  ends:  W  = —  • 


One  round  end :  W  = 


2P 
dfc2 


Af 
Both  ends  round :  W  =  - 


4/ 


Values  of  the  constants  c  and  /,  for  several  materials,  are 


Hard  steel 69,000  Ibs.  per  sq.  in. 

Structural  steel  48,000       " 

Wrought  iron 36,000       " 

Cast  iron 80,000       " 

Wood  7,200       " 

Examples: 


c 

20,000 

30,000 

36,000 

6,400 

3,000 


1.  A  hollow  cast-iron  column,  fixed  at  the  ends,  is  20  ft. 
high  and  has  a  mean  diameter  of  1  ft.     It  is  to  carry  100 
tons.     Factor  of  safety  8.     What  is  the  thickness  of  the 
metal? 

Ans.     1  in. 

2.  What  is  the  crushing  load  of  a  wrought-iron  pillar  10  ft. 
high,  3  ins.  in  diameter,  and  with  rounded  ends? 

Ans.     30  tons,  nearly. 


COLUMNS   AND  STRUTS.  Ill 

3.  What  is  the  crushing  load  of  a  cast-iron  column  having 
flat  ends,  being  15  ft.  long  and  6  ins.  in  diameter? 

Ans.     311  tons. 

4.  A  wooden  strut,  12  ft.  long,  supports  a  load  of  15  tons. 
If  the  section  is  square  what  must  be  the  side  of  the  square, 
allowing  a  factor  of  safety  of  10? 

Ans.     9.25  ins. 

5.  A  hollow  wrought-iron  column  with  flat  ends  is  20  ft. 
long,  10  ins.  outside  diameter,  7  ins.  inside  diameter.     What 
load  will  it  carry? 

Ans.     550  tons. 

6.  A  solid  steel  column  with  round  ends  is  6  ins.  in  diame- 
ter and  37  ft.  long.     What  load  will  it  bear? 

Ans.     90,000+  Ibs. 

7.  A  square  wooden  column  with  fixed  ends  is  20  ft.  long 
and  carries  a  load  of  9500  Ibs.,  with  a  factor  of  safety  of  10. 
What  is  the  side  of  the  square?     (Euler's  formula.) 

Ans.     5.769  ins. 

8.  If  in  example  2  the  pillar  were  of  rectangular  section,  of 
breadth  double  the  thickness,  what  sectional  area  would  be 
required  for  strength  equal  to  that  of  the  pillar  of  example  2? 

Ans.     9.4  sq.  ins. 


CHAPTER  XII. 

STRESS  ON  MEMBERS  OF  FRAMES. 

62.  We  have  thus  far  considered  the  strength  of  single 
pieces  of  material;  let  us  now  investigate  the  methods  for 
finding  the  stress  on  any  of  the  parts  of  a  loaded  structure 
consisting  of  several  members.  This  can  be  done  graphi- 
cally or  by  calculation,  and  both  ways  will  be  considered. 
We  will  understand  by  the  word  structure  anything  built  of 
separate  parts,  called  members,  between  which  there  is  to 
be  no  motion.  Each  member  must  be  strong  enough  to 
withstand  all  the  forces  to  which  it  will  be  subjected  with- 
out permanent  deformation. 

Forces  are  either  external  or  internal. 

External  forces  acting  on  a  structure  are  (1)  the  weights 
of  the  members;  (2)  the  loads  carried,  such  as  the  traffic 
crossing  a  bridge,  rain  and  wind  pressure  on  a  roof,  weight 
lifted  by  a  crane,  etc.;  and  (3)  the  supporting  forces. 
The  internal  forces  are  the  resistances  offered  by  the  mem- 
bers to  distortion. 

The  external  forces  acting  on  any  member  of  a  structure 
are  its  weight,  the  load  and  supporting  forces  and  the  forces 
exerted  on  it  by  other  members. 

We  shall  make  use  of  the  following  rules  for  the  equi- 
librium of  a  structure: 

(1)  The  algebraic  sum  of  all  the  external  forces  taken 
together  must  equal  zero. 

(2)  All  the  forces,  both  external  and  internal,  acting  on 
any  member  must  balance. 

(3)  All  the  external  forces  on  one  side  of  any  imaginary 

112 


STRESS  ON  MEMBERS  OF  FRAMES.  113 

complete  section  of  a  structure  must  balance  all  the  internal 
forces  on  the  same  side  of  the  section. 

(4)  The  algebraic  sum  of  the  moments  about  any  axis  of 
all  the  forces  both  external  and  internal  on  one  side  of  any 
section  through  a  structure  must  equal  zero. 

All  members  of  a  structure  are  held  together  by  joints. 

By  the  word  frame  will  be  meant  a  structure  which  has 
frictionless  joints.  Frames  cannot  actually  exist,  for  there 
is  always  frictional  resistance;  however,  many  structures 
closely  approach  them,  and  whatever  error  there  is  may  be 
considered  somewhat  as  a  factor  of  safety. 

A  member  of  a  structure  which  is  in  tension  is  called  a  tie  ; 
if  under  compression,  a  strut. 

If  two  or  more  members  of  a  frame  are  connected  at  a 
joint,  the  stress  in  the  members  is  considered  as  acting  at 
the  center  of  the  joint,  each  member  necessarily  having  a 
joint  at  each  end  to  keep  it  in  equilibrium,  and  obviously 
the  stress  in  the  member  will  be  of  the  same  amount  through- 
out and  will  act  along  the  axis,  the  effect  on  the  joint  at  one 
end  being  in  the  opposite  direction  to  the  effect  on  the  joint 
at  the  other  end. 

If  we  consider  a  single  joint  of  a  structure,  the  resultant 
of  the  stresses  in  the  members  forming  the  joint  must  be 
equal  in  amount  and  opposite  in  direction  to  the  load  on  that 
joint,  for  otherwise  the  joint  could  not  be  in  equilibrium. 

All  frames  will  be  considered  as  loaded  at  the  joints.  In 
the  case  of  continuous  loads,  such  as  the  weight  of  a  mem- 
ber, the  equivalent  parallel  forces  will  be  considered  as  act- 
ing at  the  joints;  thus  a  uniformly  loaded  member  would  be 
considered  as  having  half  the  total  load  acting  at  each  end. 

63.  As  a  simple  frame  we  will  first  consider  a  common 
triangular  roof  truss  in  which  the  weight  of  the  roof  itself 
forms  the  load,  say  w  Ibs.  per  ft.-run  on  each  rafter.  On  the 
member  AB  there  will  be  wa  Ibs.,  of  which  half  will  act  at  A 
and  half  at  B.  On  the  member  BC  there  will  be  wb  Ibs.,  of 


114 


STRENGTH   OF   MATERIAL. 


which  half  will  act  at  B  and  half  at  C;  so  that  at  B  there  will 


w 


be  a  load  of  —  (a  +  b)  Ibs. 


Now  the  supporting  forces  P 


w& 


and  Q  will  have  to  sustain  this  whole  load,  and  their  value 
is  obtained  just  as  we  found  it  in  the  case  of  loaded  beams, 
the  moment  arm  of  the  load  at  B  being  the  projection  of 
the  member  AB  on  AC,  or  BC  on  AC,  the  lengths  being 
found  from  our  knowledge  of  distances,  angles,  etc.  Now 

77V7 

the  lines  of  action  of  the  load,  —  at  A,  and  of  the  supporting 

force  P  are  identical,  but  the  directions  of  the  forces  are 

opposite,  so  we  can  use  their 
difference  for  P  without  mak- 
ing any  change  in  the  stress  of 
the  members  of  the  struc- 
ture (this  being  the  resultant 
of  the  two  forces)  and  thus 
.„.  .Q  reducing  the  number  of  ex- 

ternal forces  to  three.  (It  is 

convenient  in  calculating  the  supporting  forces  to  ignore 

the  load  which  acts  on  the  lower  end  of  the  rafters.)     We 

have    now    to   find    the 

stress    in    the    members 

AB,  BC,  and  AC,  and  we 

can  do  this  (1)  by  calcu- 
lating directly  from  our 

knowledge  of  the  loads, 

angles  and  distances;  (2) 


by  means  of  the  Ritter 
section,  and  (3)  by  a 
graphic  method.  Each 


IP 


Fig.  50. 


of  these  methods  will  be  explained  in  the  three  following  arti- 
cles, but  each  method  must  be  known  thoroughly,  for  with 
complicated  frames  it  is  necessary  to  employ  more  than  one 
of  the  methods  to  get  results.  Before  proceeding  we  will 


STRESS  ON  MEMBERS  OF  FRAMES. 


115 


introduce  a  system  of  lettering  which  will  be  convenient, 
and  which  should  be  used  in  all  cases.  Referring  to  Fig.  50, 
the  external  loads  at  the  ends  of  each  member  have  been 
indicated  by  arrow-heads.  After  each  one  of  these  exter- 
nal loads  we  place  a  capital  letter,  and  in  each  space  made 
by  the  members  of  the  frame  we  place  a  small  letter.  To 
indicate  any  force  we  write  or  mention  the  letters  on  each 
side  of  it;  for  example,  the  left  supporting  force  is  EA,  the 
right  one  DE,  the  load  at  the  peak  BC,  etc.  The  internal 
forces  will  all  have  at  least  one  small  letter,  though  both 
may  be  small;  thus  the  stress  in  the  vertical  member  from 
the  peak  down  is  be,  that  in  the  member  inclined  to  the  left 
from  the  peak  is  Bb,  the  one  inclined  to  the  right  Cc,  etc. 
With  this  method  we  letter  the  forces,  which  will  be  found 
most  convenient  when  using  the  graphic  method  for  finding 
their  values. 

64.   As  an  illustration,  we  will   find   the  stresses  in  the 
members  of  a  triangular  roof  truss  of  span  25  ft.,  rafters  of 


1120 


20  and  15  ft.  respectively,  and  which  has  a  load  of  100  Ibs. 
per  ft. -run  on  the  rafters.  The  loads,  additional  dimen- 
sions and  supporting  forces  as  indicated  in  Fig.  51  are 
found  by  the  methods  of  the  preceding  article, 

cos  0  =  sin  <f>  =  ft  =  |, 
sin  6  =  cos  <>  =     -  =    . 


116  STRENGTH  OF  MATERIAL. 

For  the  equilibrium  of  the  joint  at  the  peak  the  sum  of  the 
vertical  components  of  the  stresses  Aa  and  Ba  must  be 
equal  to  the  load  1750,  but  must  act  in  the  opposite  direc- 
tion; therefore, 

Aa  cos  cj)  +  Ba  cos  0  =  1750,  or,  f  Aa  +  f  Ba  =  1750.      (1) 

Considering  the  joint  at  the  right  support,  the  sum  of  the 
horizontal  and  of  the  vertical  components  of  the  forces 
acting  there  must  be  zero,  or  as  before  the  vertical  compo- 
nent of  Ba  must  equal  Q,  and  the  horizontal  component 
must  equal  the  stress  Ca.  Resolving,  horizontally, 

Ca  —  Ba  cos  (f>,  or,  Ca  =  \  Ba.  (2) 

Vertically, 

Ba  sin  </>  =  1120,  or,  f  Ba  =  1120.    .'.  Ba  =  1400  Ibs. 

Substituting  in  (1), 

|  Aa  +  f  .  1400  =  1750.          .'.  Aa  =  1050  Ibs. 

Substituting  in  (2), 

Ca  =  |  .  1400  =  840.  /.  Ca  =  840  Ibs. 

Now  it  is  obvious  that  the  stresses  Aa  and  Ba  will  be  com- 
pressive,  and  that  Ca  will  be  tensile,  but  in  many  cases  it  is 
not  obvious,  and  with  the  graphic  method  will  be  shown  a 
way  of  ^accurately  distinguishing.  Taking  the  joint  at  the 
peak,  the  external  load  acts  down,  so  the  resultant  of  the 
stresses  Aa  and  Ba  must  act  up;  to  do  this  they  must  push 
on  the  joint;  a  stress  then  which  pushes  on  a  joint  is  com- 
pressive  and  one  which  pulls  is  tensile. 

This  method  of  finding  stresses  will  hereafter  be  called 
the  method  by  calculation,  to  distinguish  it  from  Hitter's 
method. 

65.  Hitter's  Method.  —  This  may  be  called  the  method 
of  sections,  as  it  involves  our  fourth  rule  for  the  equilibrium 


STRESS  ON   MEMBERS  OF  FRAMES.  117 

of  a  structure.  The  moments  may  be  taken  about  any  axis, 
and  if  possible  we  choose  an  axis  about  which  the  moments 
of  all  the  unknown  stresses  cut  by  the  section  except  one 
disappear.  We  will  solve  the  problem  of  the  preceding  arti- 
cle by  this  method  and  to  do  so  will  first  consider  a  section 


Q=1120 

Fig.  52. 

HH  as  shown  in  Fig.  52.  If  now  we  take  moments  about 
the  joint  at  the  peak,  the  moments  of  the  stresses  Aa  and  Ba 
will  be  zero  and  we  will  have  to  the  left  of  the  section  only  the 
external  force  P  and  the  stress  Ca  to  deal  with;  or  to  the 
right  the  external  force  Q  and  the  stress  Ca,  there  being  no 
moment  of  the  1750-lb.  load  about  the  joint  at  the  peak. 
Taking  moments,  then,  the  perpendicular  distance  from  the 
peak  to  the  stress  Ca  being  12  ft.,  and  that  to  the  line  of 
action  of  the  force  P  being  16  ft.,  we  have 

630  X  16  -  Ca  X  12,  or,  Ca  =  840,  as  before. 

Taking  moments  about  the  joint  at  the  right  support  will 
eliminate  the  stress  Ca,  and  still  working  with  the  forces  to 
the  left  of  the  section  we  have 

630  X  25  =  Aa  X  15,  or,  Aa  =  1050,  as  before. 

Care  must  be  taken  to  use  the  perpendicular  distance  to  the 
line  of  action  of  the  forces.  We  have  used  15  in  this  case 


118  STRENGTH  OF  MATERIAL. 

because,  it  will  be  noticed,  this  frame  is  right-angled  at 
the  peak. 

To  get  the  stress  in  Ba  we  must  use  another  section,  for  it 
is  clear  that  the  only  internal  stresses  involved  are  those  of 
the  members  through  which  the  section  passes.  We  will  use 
the  section  KK  and  take  the  moments  about  an  axis  through, 
let  us  say,  the  point  L.  (We  may  use  any  point.)  Here 
the  perpendicular  distance  to  the  stress  Ba  is  7.2  ft.,  and  to 
the  supporting  force  Q  it  is  9  ft.;  so 

1120  X  9  =  Ba  X  7.2,  or,  Ba  =  1400,  as  before. 

Instead  of  choosing  L  for  the  axis  of  moments,  we  could 
just  as  well  have  taken  the  joint  at  the  left  supporting  force 
from  which  the  moments  are 

1120  X  25  =  Ba  X  20,  or,  Ba  =  1400. 

This  method  will  be  found  convenient  where  the  stress  in 
particular  members  is  desired,  and  is  frequently  necessary  in 
the  graphic  method. 

66.  The  graphic  method,  invented  by  Clerk  Maxwell,  is 
based  upon  the  principle  of  mechanics  that  a  number  of  forces 
acting  at  a  point  are  in  equilibrium  only  when  they  can  be 
represented  in  amount  and  direction  by  the  consecutive  sides 
of  a  closed  polygon.  In  all  frames  the  forces  acting  at  the 
joints  must  be  in  equilibrium,  and  the  line  of  action  of 
the  internal  forces  must  be  in  the  direction  of  the  axis  of 
the  member  in  which  they  are  found.  All  the  external  forces 
taken  in  order  around  the  frame  will  form  a  closed  polygon, 
because  by  our  first  rule  for  the  equilibrium  of  structures 
they  must  balance.  Draw  carefully  a  diagram  of  the  frame 
and  at  each  joint  indicate  by  an  arrow-head  the  load,  if  there 
be  one;  indicate  also  the  supporting  forces.  This  figure  is 
called  the  frame  diagram,  and  will  be  denoted  by  F.  D. 
The  diagram  representing  the  polygons  of  forces  acting  at 


STRESS   ON   MEMBERS   OF   FRAMES. 


119 


the  joints  is  called  the  reciprocal  diagram,  and  it  will  be  de- 
noted by  R.  D.  Having  determined  the  amount  and  direc- 
tion of  the  external  forces,  plot  them  to  a  convenient  scale, 
forming  the  external  force  polygon.  In  case  the  loads  are  all 
vertical,  as  in  the  example  of  Art.  63,  the  external  force 
polygon  is  a  vertical  straight  line,  as  in  Fig.  53,  the  load 


R.D. 


c 
F.D. 


Q=1120 


Fig.  53. 


i_LB 


1750  Ibs.  being  represented  by  AB,  the  supporting  force  Q, 
acting  up,  by  BC,  and  P  by  CA,  all  of  which  forces  have  the 
same  distinguishing  letters  in  the  R.  D.  as  in  the  F.  D.  Now 
the  stress  in  the  left-hand  rafter  is  in  the  direction  of  the 
rafter  itself,  so  from  A  of  the  external  force  polygon  draw  a 
line  parallel  to  this  rafter  to  represent  the  line  of  action  of 
the  stress  Aa,  and  from  B  draw  a  line  parallel  to  the  right- 
hand  rafter  to  represent  the  line  of  action  of  the  stress  Ba. 
The  intersection  of  these  two  lines  will  be  the  point  a,  and 
if  we  connect  C  and  a,  the  line  will  be  found  to  be  parallel 
to  the  tie  rod  of  the  frame,  and  Ca  will  represent  the  stress 
in  the  tie  rod  to  the  same  scale  as  that  used  to  plot  the  ex- 
ternal force  polygon,  just  as  Aa  and  Ba  will  represent  the 
stress  in  the  rafters.  This  figure  is  called  the  Reciprocal 


120  STRENGTH  OF   MATERIAL. 

Diagram,  and  that  the  lines  do  represent  the  stresses  may  be 
proved,  for  the  angles  BAa  and  ABa  are  equal  respectively 
to  <j>  and  6  of  the  F.  D.  by  construction,  and  as  AC  is  to  scale 
equal  to  630,  Aa  will  equal  630  sec  <£  =  630  X  f  =  1050; 
Ba  will  equal  1120  sec  0  -  1120  X  I  =1400;  and  Ca  will 
equal  630  tan  </>,  or  1120  tan  6,  either  of  which  will  give  840. 
The  results  are  then  the  same  as  those  obtained  by  other 
methods.  Of  course,  if  the  diagram  is  drawn  accurately,  the 
stresses  may  be  measured  off  to  scale. 

To  find  the  kind  of  stress  in  any  member,  notice  that  the 
forces  acting  on  the  joint  at  the  peak  are  the  load  A B  and 
the  stresses  Aa  and  Ba.  If  we  pick  these  lines  out  on  the 
R.  D.,  they  will  form  the  polygon  of  forces  (in  this  case  a 
triangle)  for  the  joint  at  the  peak.  Knowing  the  direction 
of  one  of  these  forces,  we  will  indicate  it  by  an  arrowhead; 
for  example,  we  know  AB  acts  down,  so  we  put  an  arrowhead 
pointing  down  on  the  line  AB  of  the  R.  D.  For  the  equi- 
librium of  the  joint  at  the  peak  the  direction  of  the  forces 
acting  on  it  will  be  indicated  if  we  suppose  the  arrowhead 
of  AB  to  move  in  order  around  the  sides  of  the  polygon  for 
this  joint,  starting  in  the  direction  in  which  it  points.  When 
on  each  side  it  will  indicate  the  direction  in  which  the  stress 
for  the  corresponding  member  acts  on  the  joint,  and  remem- 
bering that  a  push  indicates  compression  and  a  pull  tension, 
we  can  at  once  state  what  kind  of  stress  exists  in  any  member 
acting  on  the  joint.  The  arrowheads  are  marked  in  the  R.  D. 
of  Fig.  53  for  the  joint  at  the  peak.  They  have  been  indi- 
cated for  a  single  joint  only,  because  if  we  consider  any  other 
joint  involving  the  stress  of  any  of  the  members  acting  on  this 
one  we  would  have  two  arrowheads  on  the  same  line  pointing 
in  opposite  directions.  This  at  first  glance  appears  to  be  in- 
correct, but  when  we  remember  that  the  stress  in  any  member 
acts  in  opposite  directions  on  the  joints  at  its  two  ends,  and 
that  we  would  n<ft£  be  working  .with  a  different  joint,  the 
apparent  inaccuracy  clears  itself  up;  for  example,  if  we  con- 


STRESS   ON   MEMBERS   OF   FRAMES.  121 

sider  the  joint  at  the  right-hand  support,  we  will  find  the 
arrowhead  for  the  stress  Ba  pointing  toward  B  in  the  R.  D., 
which  we  know  to  be  correct,  as  the  right-hand  rafter  is  in 
compression  and  therefore  pushes  on  this  joint. 


Examples  : 

1.  The  slope  of  the  rafters  of  a  simple  triangular  roof 
truss  is  30°.     What  is  the  stress  in  each  member  when  loaded 
with  250  Ibs.  at  the  peak? 

Ans.     Rafters,  250  Ibs.;  tie  rod,  216.5  Ibs. 

2.  A  beam  15  ft.  long  is  trussed  with  steel  tension  rods 
and  a  strut  at  the  middle  forming  a  simple  triangular  truss 
2  ft.  deep.     What  is  the  stress  on  each  member  when  loaded 
with  2  tons  at  the  middle? 

Ans.     Strut,  2  tons;  tension  rods,  3.88  tons;  thrust  on  the 
beam,  3.75  tons. 

3.  A  small  brow,  6  ft.  broad  and  20-ft.  span,  carries  a  load 
of  100  Ibs.  per  sq.  ft.  of  platform.     It  is  supported  by  two 
simple  triangular  trusses  3  ft.  deep.     Find  the  stress  in  each 
member. 

Ans.     Strut,  3000  Ibs.;  tie  rods,  5220;  thrust  on  beams, 
5000  Ibs. 

4.  The  rafters  of  a  simple  triangular  roof-truss  slope  30° 
and  45°;  span  10  ft.;  the  rafters  are  2^  ft.  apart,  and  the 
roof  weighs  20  Ibs.   per  sq.   ft.     Find  the  stress  on  each 
member. 

Ans.     Stress  on  tie  rod,  198  Ibs. 

5.  A  small  brow,  like  that  of  example  3,  4  ft.  broad  and 
20-ft.  span,  carries  a  load  of  100  Ibs.  per  sq.  ft.  of  platform. 
A  load  of  1  ton  passes  over  the  brow;  what  is  the  stress  in  the 
members  when  this  load  is  in  the  middle  of  the  brow? 

Ans.     Compression  of  strut,  3120  Ibs. 


122  STRENGTH  OF  MATERIAL. 

6.  The  tie  rod  of  a  simple  triangular  roof-truss  is  25  ft. 
long  and  inclined  at  30°  to  the  horizontal.  The  supports  are 
at  its  ends  and  the  rafters  from  its  ends  are  20  and  15  ft. 
long  and  are  loaded  with  30  Ibs.  per  ft.-run.  What  is  the 
stress  in  each  member? 

Ans.  Stress  in  rafters,  148+  Ibs.  and  512  +  Ibs.;  tie 
rod,  174+  Ibs. 


CHAPTER   XIII. 

FRAMED  STRUCTURES  —  Continued. 

67.  A  roof  truss  with  a  vertical  member  from  peak  to  tie 
has  flooring  laid  on  the  horizontal  ties,  in  addition  to  the 
load  on  the  rafters,  so  that  when  divided  at  the  joints  the 
load  will  be  as  shown  in  Fig.  54. 


R.D. 


Fig.  54. 

^ 

Proceeding  in  the  usual  way  we  get  the  R.  D.  as  shown, 
and  from  it 

Aa  =  10  tons  C.   (compression). 
Bb  =  8  \/2  tons  C. 
ba  —  3%  tons  T.   (tension). 
Da  =  Cb  =  8  tons  T. 

To  get  the  stress  Aa,  for  example,  by  the  method  of  sections; 
taking  a  section  HH,  and  moments  about  the  joint  at  the 
right-hand  support,  using  forces  to  the  left  of  the  section, 
we  get,  x  being  equal  to  16.8  ft., 

P  X  28  =  Aa  X  x,  or,  Aa  =  10  tons,  as  before, 
123 


124 


STRENGTH  OF  MATERIAL. 


The  method  by  calculation  is  seldom  used,  as  compared  to  the 
other  methods  it  is  complicated. 

68.    A  "  King-Post  Truss,"  slope  of  rafters  45°,  and  having 
struts  to  the  middle  points  of  the  rafters  as  shown  in  Fig.  55, 


E 

F.D. 


QL. 

Fig.  66. 


has  a  load  of  8  tons  uniformly  distributed  on  each  rafter. 
Find  the  stress  in  the  members.  The  loads  are  as  shown  in 
the  F.  D.  The  R.  D.  gives  us: 

Aa  =  Dd  =6  Wtons  C. 
Bb  =  Cc  =4\/2^tonsC. 
ab  =  cd   =  2  V2  tons  C. 
Ea  =  Ed  =  6  tons  T. 
Be  =  4  tons  T. 

To  get  the  stress  Ed,  for  example,  by  method  of  sections, 
take  section  HH,  moments  about  joint  at  peak,  calling 
span  S. 

S  S 

Q  X  -  =  Ed  X  -  •      . '.  Ed  =  6  tons,  as  above. 
2  2 

69.  In  the  preceding  example,  suppose  the  roof  had  also  to 
sustain  a  horizontal  wind  pressure  of  6  tons,  uniformly  dis- 
tributed on  the  right-hand  rafter.  Putting  in  all  the  loads 
they  would  be  as  shown  in  Fig.  56,  and  the  walls  would  have 
to  sustain  in  addition  to  the  vertical  load  a  lateral  pressure 
of  6  tons.  The  resultant  of  the  loads  is  not  now  vertical, 


FRAMED  STRUCTURES. 


125 


therefore  our  supporting  forces  will  be  inclined.  Obviously, 
the  effect  is  greatest  on  the  left-hand  support,  so  the  inclina- 
tion of  P  will  not  be  the  same  as  that  of  Q.  Calling  the 


R.D. 


Fig.  56. 

angles  that  P  and  Q  make  with  the  vertical  (j>  and  6  re- 
spectively, and  the  span  S,  and  taking  moments  about  the 
joint  at  the  left  support  (Fig.  56,  a), 

Sf 

QScos  #  =- 
4 


S>  Sf  Sf 

-X4-f-f£  X4  +  2S--X  3  -  -X  1.5. 
2  42 


.-.  Q  cos  6=  6.5, 
and,  by  the  same  method, 

P  cos  $=  9.5, 

which  shows  the  sum  of  the  vertical  components  of  the  sup- 
porting forces  to  be  equal  to  the  vertical  loads  as  it  should 
be,  also 

P  sin  <j)  +  Q  sin  0  =  6  tons, 


126  STRENGTH  OF  MATERIAL. 

\ 

or  the  horizontal  components  equal  the  horizontal  loads.  Pro- 
ceed now  with  the  external  force  polygon  to  and  including 
the  force  HI.  The  next  two  forces  are  the  supporting  forces. 
We  know  the  external  forces  give  us  a  closed  polygon,  there- 
fore the  resultant  of  the  two  supporting  forces  is  the  dotted 
line  I  A.  We  know  the  sum  of  the  horizontal  components  of 
the  two  supporting  forces  is  6  tons,  so  we  will  draw  a  line 
parallel  to  the  direction  of  AB,  BC,  etc.,  at  a  distance  from  it 
equal  to  6  tons  as  per  our  scale.  We  also  know  the  vertical 
component  of  the  force  P  is  9.5  tons,  so  laying  off  from  A 
toward  D  the  distance  9.5  per  scale,  if  through  this  point  we 
draw  a  horizontal  line,  the  point  J  must  be  somewhere  on  it, 
and  also  somewhere  on  the  first  line.  Therefore  J  is  located 
at  their  intersection.  Connecting  J  and  /  we  find  the  line  is 
vertical  and  therefore  that  the  supporting  lorce  at  Q  is  ver- 
tical and  equal  to  6.5  tons.  Connecting  J  and  A  we  have  the 
supporting  force  P  both  in  magnitude  and  direction,  and  find 
it  equal  to  11.236  tons  and  at  32°  16'  32"  with  the  vertical. 

Of  course  practically  the  right-hand  support  would  sustain 
some  of  the  horizontal  load,  due  to'  friction  and  the  way  the 
roof  is  secured  to  the  walls,  but  if  the  left-hand  wall  is  built 
strong  enough  to  take  all  the  horizontal  effect  our  structure 
will  be  safe.  As  we  can  always  resolve  any  oblique  force  hori- 
zontally and  vertically,  we  may  assume  that  the  supporting 
wall,  on  the  side  from  which  the  horizontal  component  comes, 
has  a  roller  on  its  top,  so  that  the  supporting  force  on  that 
side  will  have  to  be  vertical. 

Proceeding  now,  we  complete  the  F.  D.  by  putting  in  the 
supporting  forces,  then  finish  the  R.  D.  as  shown  in  the  figure 
and  find  the  stresses  to  be  as  follows: 

Ba  =  7.5  \/2~tons  C.  EC  =  4  V2~tons  C. 
Cb  =  5.5  V:2  tons  C.  be  =  5.5  tons  T. 

Gd  =  4.5  \/2"tons  C.  cd  =  3.5  \/2  tons  C. 

ba  =  2  \/2~tons  C.  Jd  =  3  tons  T. 
Ja  =  1.5  tons  T. 


FRAMED  STRUCTURES. 


127 


To  get  the  stress  cd  by  the  method  of  sections,  we  must  know 
the  stress  Jd,  then  take  a  section  KK  and  use  the  forces  to 
the  right  of  the  section  taking  moments  about  the  joint  at  the 
peak.  For  Jd,  using  the  section  HH, 

o  o  rr  O 

«^X-=6.5X  '--2X--  1.5  X-, 

or, 

Jd  =  3  tons  .  T. ; 

now,  using  section  KK  (a  section  through  EC,  cd  and  dJ), 


cd  X  =-  =  Jd  X  - 

2  V2  2 


S  S 

L5X2~6<52 


S 


S 
-, 


from  which  cd  =  3.5  \/2  .  C.,  as  before. 

70.  An  inverted  "  Queen-Truss  "  is  loaded,  as  shown  in 
the  F.  D.  of  Fig.  57,  with  unequal  loads  over  the  two  struts. 
With  this  loading  the  R.  D.  is  as  shown.  The  stress  in  the 
diagonal  is  tension.  If  the  loads  had  been  reversed,  i.e., 2  W 


R.D. 


Fig.  57. 


for  BC,  and  W  for  AB,  the  stress  in  the  diagonal  would 
have  been  compressive.  If  the  diagonal  member  had  been 
omitted,  the  R.  D.  would  not  have  been  a  closed  figure 


128 


STRENGTH  OF   MATERIAL. 


(except  in  the  case  of  equal  loads  over  the  struts),  showing 
that  the  frame  would  not  then  be  in  equilibrium.  The 
diagonal  member  is  then  necessary  in  bridges  of  this  kind, 
because  of  the  traveling  loads  they  must  carry. 

71.  A  Bollman  truss  is  24  ft.  long,  3  ft.  deep  and  carries 
a  uniform  load  of  3  tons  per  ft.-run. 

The  F.  D.  is  as  shown  in  Fig.  58.  Proceeding  with  the 
R.  D.  we  draw  the  external  force  polygon,  the  direction  of 


24  tons  /        24  tons 

I       '  i 

A        I       <B         I        C 


F.D. 


B.D. 


R,D, 


Fig.  58. 


the  stresses  Aa,  Bb,  Cf,  Dg,  De,  DC,  and  Db.  Here  we  have 
to  stop,  for  we  have  none  of  the  points  represented  by  the 
small  letters.  The  simplest  way  out  of  our  difficulty  will  be 
to  find  the  amount  of  stress  in  one  of  the  members  by  the 
method  of  sections.  Use  the  section  HH,  taking  moments 
about  the  point  where  the  two  long  tension  members  cross 
(marked  with  circle)  and  working  with  the  forces  to  the  left 
of  the  section, 

Bd  X  |  =  24  X  12  -  24  X  4,  or,  Bd  =  85J  tons. 

Laying  this  value  off  to  scale  on  the  line  Bd  gives  us  the 
point  d,  after  which  we  have  no  difficulty  in  obtaining  the 
R.  D.  shown. 

72.  The  force  diagram  of  Fig.  59  shows  an  N  girder  of 
four  panels  with  a  uniform  load  over  the  lower  platform. 
Having  drawn  the  external  force  diagram,  the  stress  Ea 
being  horizontal,  the  point  a  must  lie  somewhere  on  a  hori- 


FRAMED  STRUCTURES. 


129 


zontal  line  through  E;  the  stress  Aa  being  vertical  the  point 
a  must  lie  somewhere  on  a  vertical  line  through  A.  There- 
fore the  point  a  coincides  with  the  point  E  of  the  R.  D.,  or 
the  stress  Ea  is  equal  to  zero  with  this  loading.  The  mem- 
ber is  necessary  to  the  bridge,  however,  for  even  if  it  did 
not  support* part  of  the  platform  it  would  be  required  to 
keep  the  frame  from  turning  about  the  upper  joint  over 
the  left  support.  The  stress  Bh  also  proves  to  be  equal  to 
zero  in  the  same  way.  Proceeding  now  we  get  the  R.  D.  as 
shown  and  finding  the  stress  ed  is  also  zero.  In  this  case, 


E         4.          D        4.         C         |          B 
WWW 

F.D. 


R.D. 


Fig.  59. 


too,  the  member  is  necessary,  for  with  a  joint  at  the  middle 
of  the  top  boom  the  stresses  Ad  and  Ae  being  compressive 
would  cause  it  to  turn,  and  without  the  joint  the  necessary 
length  of  the  member  would  cause  it  to  buckle.  Having 
now  the  R.  D.  the  stresses  in  the  members  are  easily 
obtained.  In  designing  N  girders  we  must  consider  the 
effect  of  traveling  loads,  for  the  bridge  in  addition  to  sus- 
taining its  own  weight  must  be  strong  enough  to  sustain 
the  stresses  caused  by  the  loads  which  cross  it.  The  method 
of  constructing  bridge  platforms  (the  planks  or  railroad 
ties  being  placed  across  stringers  between  joints)  puts  prac- 
tically all  the  bending  stresses  on  the  horizontal  parts  and 
all  the  shearing  stresses  are  sustained  by  the  diagonals  (the 
shearing  stress  being  equal  to  the  vertical  component  of 
the  stress  in  the  diagonal).  Therefore,  considering  only  the 


130  STRENGTH  OF   MATERIAL. 

weight  of  the  structure,  the  diagonals  at  the  left  of  the 
middle  are  inclined  to  the  left  and  those  to  the  right  of  the 
middle  to  the  right,  because  those  directions  put  them  in 
tension.  (For  the  same  load  a  rod  in  tension  requires  less 
cross-sectional  area  than  one  in  compression,  hence  less 
material,  less  weight  and  less  cost.)  If  we  consider  a  bridge 
of  this  kind  with  a  concentrated  load  moving  across  it, 
referring  to  Fig.  41,  Art.  50,  the  load  entering  the  bridge  at 
the  left  end,  we  see  that  positive  shearing  force  predomi- 
nates at  the  left  end  and  negative  at  the  right,  the  curve  in 
this  figure  being  for  the  traveling  load  only.  Therefore  the 
total  shearing  force  due  to  both  the  weight  of  the  structure 
and  the  traveling  load  will  change  sign  from  plus  to  minus 
at  some  position  of  the  traveling  load  other  than  the  middle 
of  the  bridge.  Obviously,  this  wrill  change  the  kind  of  stress 
in  the  diagonal  at  this  point,  and  the  dimensions  of  the 
diagonal  here  must  be  made  such  as  to  sustain  the  new 
stress,  or  we  must  put  in  a  second  diagonal  inclined  in  the 
opposite  direction.  The  latter  method  is  used  because  of 
the  increase  of  weight  necess-ary  with  a  diagonal  under  com- 
pression, and  also  because  a  tension  rod  is  less  expensive 
and  quite  as  efficient.  These  extra  diagonals  are  called 
counter  braces,  and  the  above  is  the  reason  for  counter- 
bracing  some  of  the  middle  panels  of  N  girders. 

ff_ 


\A 


c 
R.D. 


Fig.  60. 

73.   A  Warren  girder,  angle  between  members  60°,  carries 
uniform  load  on  the  lower  platform  as  shown  in  Fig.  60, 


FRAMED  STRUCTURES.  131 

The  R.  D.  is  as  shown,  and  the  stresses  in  members  are 
easily  obtained  from  it.  The  remarks  of  the  preceding 
article  on  counter  bracing  apply  equally  to  this  girder.  If 
we  draw  vertical  lines  through  each  joint  and  call  the  part 
between  two  vertical  lines  a  panel,  the  calculation  will  be 
just  the  same  as  for  the  N  girder. 

Examples: 

1.  A  King- Post  Truss,  slope  of  rafters  45°,  has  struts  to 
the  middle  point  of  the  rafters  as  in  Fig.  55.      AB=  2  w\ 
BC  =  3  w,   and  CD=  4  w.     Find  the  stress  in  each  member 
by  the  method  of  sections  and  also  from  the  R.  D.,  stating 
the  kind. 

2.  A  bridge  is  constructed  of  a  pair  of  Warren  girders 
with  the  platform  on  the  lower  boom,  which  is  of  four  divi- 
sions.    Forty  tons  are  uniformly  distributed  over  the  left  half 
of  the  bridge.     Find  the  stress  in  each  member,  stating  the 
kind. 

3.  A  bridge,  60-ft.   span,  is  supported  by   a  pair  of  N 
girders  of  6  panels,  height  8  ft.    The  platform  is  on  the  upper 
boom  and  carries  a  uniformly  distributed  load  of  12  tons  on 
the  left  half  of  the  bridge.     Find  stress  in  each  member, 
stating  the  kind. 

4.  A  roof  is  constructed  of  two  rafters  at  right  angles;  a 
horizontal  member  connects  the  middle  points  of  the  two 
rafters;  the  ends  of  this  member  are  connected  with  lower 
ends  of  the  opposite  rafter.     The  joints  of  the  peak  and  the 
middle  of  each  rafter  carry  a  load  of  1  ton.     Find  the  stress 
in  each  member  and  the  kind,  if  the  span  is  20  V2  ft. 

5.  A  .bridge  of  60-ft.  span  is  constructed  of  a  pair  of  War- 
ren girders;  the  upper  boom,  which  supports  the  platform,  is 
of  6  divisions,  the  lower  boom  of  5.     The  platform  is  also 


132  STRENGTH  OF  MATERIAL. 

supported  by  struts  from  the  joints  of  the  lower  boom,  and 
carries  a  load  of  f  ton  per  ft.-run.  The  supporting  forces 
are  at  the  ends  of  the  upper  boom.  Find  the  amount  and 
kind  of  stress  in  each  member. 

6.  A  Bollman  truss  of  48-ft.  span  and  12  ft.  deep  carries 
a  uniform  load  of  }  ton  per  ft.-run  over  the  left  two-thirds 
of  the  horizontal  boom.     Find  the  amount  and  kind  of  stress 
in  each  member  by  the  method  of  sections  and  also  from 
the  R.  D. 

7.  A  bridge,  96-ft.   span,  is  supported  by  a  pair  of  N 
girders  with  8  panels  9  ft.  deep.     The  platform  rests  on  the 
upper   boom   and   is  loaded  with  96  tons  uniformly  distrib- 
uted.    Find  the  amount  and  kind  of  stress  in  each  member. 

8.  A  concentrated  load  of  30  tons  is  to  pass  over  the 
bridge    of    example    7.     What  panels  should    be    counter- 
braced? 

9.  A  Warren   girder,   9  ft.   long,   projects   from   a  wall. 
The  top  boom  is  of  three  divisions,  the  lower  of  two,  but 
has  a  horizontal  strut  from  the   inner   joint  to   the  wall. 
With  a  load  of  2  tons  at  the  outer  end  of  the  upper  boom 
find  the  stress  in  all  the  members,  stating  the  kind. 

10.  Suppose  the  load  of  example  9  at  each  of  the  other 
two  joints  of  the  upper  boom,  and  thence  deduce  the  results 
for  a  distributed  load  of  f  ton  per  ft.-run. 


CHAPTER  XIV. 

FRAMED   STRUCTURES  —  Continued. 

74.  An  example  of  a  scissors-beam  truss  is  shown  in  Fig. 
61.  With  vertical  loads  the  R.  D.  shown  is  readily  obtained. 
With  all  roof-trusses  the  loads  due  to  wind  pressures  must 
be  taken  into  account.  We  have  seen  that  this  may  be 


Fig.  81. 


done  by  resolving  the  wind  loads  into  their  horizontal  and 
vertical  components  (Art.  69)  and  considering  the  total 
effect  as  sustained  by  one  of  the  walls.  It  may  also  be  done 
by  combining  the  loads  on  each  joint  and  using  the  result- 
ant load  on  each  joint  in  drawing  the  external  force  dia- 
gram, or  by  finding  the  R.  D.  for  the  vertical  and  inclined 
loads  separately  and  algebraically,  adding  the  stress  due  to 
each  for  the  total  stress  in  the  members. 

75.  When  the  external  loads  and  the  members  of  a  struc- 
ture are  all  in  the  same  plane  we  can  always  find  the  stress 
in  the  members  by  means  of  the  reciprocal  diagram;  for 

133 


134 


STRENGTH  OF   MATERIAL. 


example,  the  F.  D.  of  Fig.  62  shows  a  common  crane  carry- 
ing a  weight  W.  The  external  forces  are  as  shown,  the 
forces  EC  and  DA  being  obviously  necessary  to  prevent 
overturning,  their  value  being  readily  found  by  taking 


F.D. 


R.D. 


Fig.  62. 


moments.  The  force  CD  is  equal  to  the  weight.  The  ex- 
ternal force  diagram  is  a  rectangle.  The  forces  CD  and  DA 
can  be  combined  if  desired,  their  resultant  being  equal  to  the 
dotted  line  CA.  The  R.  D.  is  readily  completed  as  shown, 
and  from  it  the  stresses  are  easily  found.  Cranes  are 
usually  arranged  so  that  the  load  is  lifted  by  means  of  a 
tackle,  the  hauling  part  of  which  leads  down  to  a  drum 
secured  somewhere  along  the  crane  post.  The  stress  in 

W 
this  rope  (which  is  equal  to  — ,  where  n  is  the  number  of 


FRAMED   STRUCTURES. 


135 


parts  of  rope  between  the  two  blocks  of  the  tackle)  aug- 
ments the  compression  in  the  jib  and  reduces  or  increases 
the  tension  of  the  jib  stay,  depending  on  whether  the  drum 
is  above  or  below  the  point  where  the  jib  joins  the  crane 
post.  These  additional  loads  on  the  members  must  be 
taken  into  account.  This  can  be  done  by  introducing  an 


Fig.  63. 


additional  external  load  at  the  end  of  the  jib,  acting  in  the 

W 

direction  of  the  hauling  part  of  the  tackle  and  equal  to 

n 

This  additional  load  will  cause  the  supporting  force  DA  of 
Fig.  63  to  be  somewhat  inclined. 

To  draw  any  external  force  diagram  we  must  know  all 
but  one  of  the  external  forces.  In  this  case  we  know  only 
the  weight  lifted  and  the  tension  of  the  rope.  We  can, 


136  STRENGTH  OF   MATERIAL. 

however,  find  the  force  CD  by  taking  moments  about  the 
lower  end  of  the  crane  post,  as 

W 

W  X-x  =     -X  z  +  CD  X  y, 
n 

whence 

W  (nx  -  z) 


CD 


ny 


The  force  DA  is  the  resultant  of  a  horizontal  force  similar 
to  DA  of  Fig.  62  and  the  slightly  inclined  supporting  force. 
Except  the  resultant  now  we  know  all  the  external  forces, 
so  our  external  force  polygon  is  A  BCD,  the  line  DA  repre- 
senting the  amount  and  direction  of  the  supporting  force. 
The  R.  D.  is  now  readily  found  to  be  as  shown,  and  the 
stresses  are  easily  obtained. 

76.  Incomplete  Frames.  —  A  frame  having  just  enough 
members  to  enable  it  to  retain  its  shape  under  any  kind  of 
loading  is  complete.  Frames  may  have  more  members  than 
necessary,  under  which  circumstances  some  of  the  members 
are  redundant,  but  frames  without  a  sufficient  number  of 
members  to  enable  them  to  retain  their  shape  under  all 
circumstances,  with  any  kind  of  loading,  are  said  to  be 
incomplete.  When  frames  are  incomplete  we  will  find  the 
reciprocal  diagram  for  different  loadings  giving  intersec- 
tions for  the  same  point  in  two  or  more  places;  as,  for 
example,  we  know  the  mansard  roof  truss,  shown  in  the 
F.  D.  of  Fig.  64,  is  incomplete  because  there  is  nothing  to 
prevent  the  three  upper  joints  from  turning.  Suppose 
Wl  =  W3  then  drawing  the  R.  D.,  we  find  the  point  a  falls 
in  two  different  places  on  the  horizontal  line  from  E]  this 
is  clearly  imposible,  for  the  stress  Ea  cannot  have  two 
values.  The  frame  then  is  not  in  equilibrium.  If  Wl  had 
not  been  equal  to  W%  the  two  a's  would  not  have  been  on 
the  same  horizontal  line.  We  could  arrange  the  loads  so 


FRAMED  STRUCTURES. 


137 


that  this  frame  would  appear  from  the  R.  D.  to  be  com- 
plete; as,  assume  either  of  the  points  a  to  be  correct, 
say  the  right  one,  and  from  it  draw  lines  (dotted  in  figure) 


E 

F.D. 


Jt.D. 


Fig.  64. 


R.D. 


Fig.  65. 


parallel  to  the  stresses  Ba  and  Ca.  These  lines  will  inter- 
sect the  external  force  diagram  at  points  indicating  loads 
which  would  put  the  frame  in  equilibrium,  but  the  first  puff 
of  wind  would  change  the  loading  and  cause  the  frame  to 
collapse. 


138  STRENGTH  OF   MATERIAL. 

To  make  this  frame  complete,  at  least  two  additional 
members  are  necessary,  for  there  are  three  joints  which  are 
likely  to  turn. 

The  two  additional  members,  shown  in  the  F.  D.  of  Fig. 
65,  will  complete  the  frame,  as  is  shown  by  the  R.  D.  being 
a  closed  figure  and  possessing  no  duplicate  points,  no  matter 
what  loads  are  applied. 

Neither  of  these  additional  members  alone  will  do  this,  as 
will  be  made  clear  if  we  discard  one  of  them,  vary  the  loads 
and  draw  the  R.  D.'s. 

A  frame,  then,  to  be  complete,  must  remain  in  equi- 
librium under  any  loads  which  do  not  stress  its  members 
beyond  the  elastic  limit. 

77.  We  have  been  considering  in  all  these  structures  the 
stress  which  acts  along  the  axis  of  the  member.  When 
members  are  not  straight,  or  when  they  carry  loads  at 
points  other  than  the  joints,  the  stress  on  them  is  not 
simply  a  thrust  or  pull  in  the  direction  of  the  axis,  but 


includes  a  bending  and  a  shearing  action.  Fig.  66  shows  a 
single  member  of  a  structure  with  two  intermediate  verti- 
cal loads.  If  we  resolve  all  these  forces  along  and  perpen- 
dicular to  the  member,  the  components  perpendicular  to 
the  member  will  cause  bending  and  shearing  stresses  in  it, 
and  using  these  components,  W  cos  0  and  Wl  cos  6,  as  the 


FRAMED  STRUCTURES.  139 

loads,  and  P  cos  6  and  Q  cos  0  as  the  supporting  forces,  we 
find  the  shearing  and  bending  action  just  as  we  did  in  Chap- 
ters VI  and  VII. 

These  intermediate  loads  cause  also  a  stress  along  the 
axis  of  the  member,  and  this  stress  (the  thrust  T  shown) 
between  W  and  W^  will  be  equal  to  Wl  sin  d,  and  between  P 
and  W  it  will  be  the  sum  of  Wt  sin  0  and  W  sin  6,  or 
(W  +  TF,)  sin  0,  which  value  is  the  reaction  of  this  member 
on  the  one  next  below  it. 

Obviously  (P  +  Q)  sin  0  causes  an  equal  reaction  on  the 
member  next  above  it. 

Just  here  we  can  show  that  our  assumption  that  the  loads 
act  on  the  joints  is  correct,  for  the  action  on  these  joints  of 
the  structure,  considered  as  a  whole,  caused  by  this  mem- 
ber is  equal  to  a  force  P  acting  downward  at  the  left  end 
and  a  force  Q  acting  downward  at  the  right  end;  and  obvi- 
ously without  error  we  can  use  these  forces  P  and  Q 
(reversed)  instead  of  considering  the  separate  loads.  When 
a  member  carries  a  uniform  load,  as  is  usually  the  case,  the 
values  of  P  and  Q  are  each  half  the  load,  and  the  thrust  T 
is  the  value  of  the  axial  stress  at  the  middle  of  the  member. 
We  have  not  considered  the  bending  and  shearing  stresses 
in  members  of  a  structure  until  now,  because  they  are  com- 
paratively small.  A.  compression  member,  being  necessarily 
large  to  sustain  the  thrust  due  to  all  the  other  members 
which  affect  it,  will  be  little  inconvenienced  by  the  small 
stress  due  to  the  bending  and  shearing  caused  by  its  own 
load,  and  the  tension  members,  in  properly  built  structures, 
will  have  only  their  own  weight  to  cause  bending  and 
shearing. 

Examples: 

1.  A  distributed  load  of  4  tons  is  carried  by  a  scissors- 
truss  whose  rafters,  12  ft.  long,  are  at  an  angle  of  90°. 
Two  of  the  three  other  members  join  the  lower  end  of  one 


140  STRENGTH   OF  MATERIAL. 

rafter  with  the  middle  of  the  other.  The  third  is  horizontal 
and  joins  the  middle  points  of  the  rafters.  Find  the  stress 
in  all  the  members. 

2.  In  example  4,  Chapter  XII,  find  the  shearing  force  and 
bending  moment  and  thrust  for  each  point  of  each  rafter 
and  draw  curves  showing  results. 

3.  A  simple  triangular  frame,  sides  3,  4  and  5  ft.  long,  has 
the  5-ft.   side  horizontal.     Its  members  weigh   10  Ibs.   per 
ft.-run,   and  the  inclined   sides  each  have   50  Ibs.   at   the 
middle.     Draw  curves  of  thrust,  shearing  force  and  bend- 
ing moment  for  each  member. 

4.  A   suspension   bridge   carries   a   plaftorm   8   ft.   wide, 
span  63  ft.  suspended  by  6  equidistant  tension  rods.     The 
lowest  joint  is  7  ft.  below  the  highest,  and  the  cable  is 
formed  of  7  straight  members.     It  carries  a  load  of  1  cwt. 
per  sq.   ft.   of  platform.     Find  the  sectional   areas  of  the 
cable  members,  allowing  a  stress  of  4  tons  per  sq.  in.  for  the 
material.     Is  this  frame  complete? 

Ans.     From  left  end,  3.72,  3.6,  3.5  sq.  in.;    middle,  3.47 
sq.  in. 

5.  A  mansard  roof  truss,  as  in  Fig.  65,  carries  1,  2  and 
3  tons  on  the  left,  top  and  right  joints.     The  rafters  make 
angles  of  45°  and  30°  with  the  horizontal.     Find  the  stress 
in  all  the  members. 

6.  A  crane,  as  in  Fig.  62,  has  the  tension  members  Ac 
horizontal  and  6  ft.  long.     The  members  be,  ba  and  aC  are 
respectively  8,  6  and  4.5  ft.  long.     Find  the  stress  in  each 
member  when  supporting  a  weight  of  12  tons. 


n  m  n 


MISCELLANEOUS   PROBLEMS. 

78,  Reinforced  Concrete  Beams.  —  Concrete  is  much  used 
in  building  at  present,  and  though  its  tensile  strength  is 
very  low  compared  to  other  materials,  the  ease  with  which 
it  is  handled,  transported  and  shaped,  together  with  the  fact 
that  when  reinforced  with  steel  its  tensile  strength  is  satisfac- 
tory and  the  structure  is  non- 
combustible,  has  made  it  most 
acceptable  to  builders.  The 
determination  of  the  constants 
for  concrete  is  rather  difficult,  N 
considering  the  changes  they 
suffer,  due  to  the  different  kinds 
of  cement  used  and  the  different 
methods  of  mixing.  Strength 

tests  for  the  cement  itself  vary  from  40  to  1000  Ibs.  per 
sq.  in.  for  tension  and  from  1100  to  12,000  Ibs.  per  sq.  in. 
for  compression. 

In  all  building  the  endeavor  has  been  to  subject  the  con- 
crete to  little  or  no  tensile  stress,  but  to  have  all  that  stress 
sustained  by  the  steel  reinforcement,  allowing  the  concrete 
to  sustain  as  much  as  possible  of  the  compressive  stress. 
Tests  have  shown  the  compressive  strength  of  concrete  to 
vary  from  750  to  5360  Ibs.  per  sq.  in.  and  the  modulus  of 
elasticity  to  range  from  500,000  to  4,167,000  in  inch-pound 
units. 

A  method  of  finding  the  moment  of  inertia  of  the  section 
of  a  reinforced  concrete  beam  is  as  follows:  Assuming  that 
the  elongation  of  the  steel  reinforcement  is  the  same  as  for 
the  concrete  at  equal  distances  from  the  neutral  plane,  which 

141 


142  STRENGTH  OF   MATERIAL. 

will  be  true  if  the  concrete  adheres  closely  to  the  metal  rein- 
forcement, we  havd  by  Hook's  law, 

P  steel  =  E  'steel  X  extension 
and 

Pconcrete  =  ^  concrete  X  extension, 

or, 


= 

' 


Es     E 
from  which 

Es 

Ps  =  TT  PC- 
EC 

Fig.  67  represents  the  section  of  a  concrete  beam  rein- 
forced by  three  steel  rods  on  the  tension  side  of  the  neutral 
plane,  N.  P.  If  now  we  suppose  the  beam  to  be  entirely  of 
concrete  and  to  retain  the  same  strength  and  the  same  depth, 
we  will  have  to  broaden  the  section  in  wake  of  the  steel  rods 
to  conform  with  the  fiber  stress  which  that  part  of  the  section 
can  now  withstand.  If  y  is  the  distance  from  the  neutral 
axis  to  an  elementary  area,  dA,  of  the  steel,  the  moment  of 
the  stress  about  the  neutral  axis  will  be 

E, 

ypsdA=ypc-fdA. 
EC 

So,  if  the  section  is  to  be  considered  homogeneous  (all  con- 
crete), we  must  have  the  areas  vary  as  the  moduli  of  elastic- 
ity, or  the  area  of  concrete  which  is  equivalent  to  the  present 
area  of  steel  will  be  as  Es  is  to  Ec.  The  equivalent  concrete 
section  will  therefore  be  something  like  that  shown  in  Fig.  68. 
The  general  equation  of  bending  gives  for  the  fiber  stress 

My 


MISCELLANEOUS  PROBLEMS.  143 

7  being  the  moment  of  inertia  of  this  equivalent  section.  For 
example,  suppose  the  beam  of  Fig.  67  to  be  6  in.  wide,  10  in. 
deep,  and  the  reinforcing  rods  to  be  \  in.  square  with  their 
lower  faces  1  in.  from  the  surface  of  the  concrete.  Let  E8 
be  to  Ec  as  15  to  1.  Here  the  area  of  the  steel  section  is 
|  sq.  in.,  hence  the  equivalent  area  of  concrete  is 

15  X  |  =  -V-  sq.  in., 

and  as  the  rods  are  half  an  inch  deep  the  additional  breadth 
at  this  part  of  the  section  due  to  the  steel  will  be  -5/  ins. 
But  there  is,  in  addition  to  the  steel,  f  ins.  of  concrete  at 


Fig.  68. 

this  part  of  the  section,  so  the  whole  width  of  the  section 
at  this  point  will  be 

V  +  I  =  ¥  ins., 

and  the  depth  of  the  broadened  part  will  be  the  same  as  the 
steel,  or  \  in.  We  have  now  all  the  dimensions  of  the  equiva- 
lent concrete  section,  from  which  we  find  that  the  neutral  axis 
is  close  to  4.4  ins.  from  the  bottom  of  the  beam,  and  the 
moment  of  inertia  of  the  section  about  this  axis  is  about  626 
in  inch-units. 

It  has  been  shown  by  experiment  that  small  cracks  appear 
on  the  tension  side  of  a  reinforced  concrete  beam  as  soon  as 
the  fiber  stress  reaches  the  limiting  tensile  strength  of  the 
concrete.  From  this  fact  it  is  clear  that  practically  all  of 
the  tensile  stress  is  supported  by  the  steel  reinforcement. 


144 


STRENGTH  OF  MATERIAL. 


6 ->! 


With  this  assumption  then,  referring  to  Fig.  69,  the  sum  of 
all  the  tensile  stresses  on  one  side  of  a  section  must  equal 

the  sum  of  all  the  compressive 
stresses  on  the  same  side  of  the 
section.  If  then  A  is  the 
known  area  of  the  steel  section 
and  Ps  the  stress  in  it;  bx  the 
area  of  the  concrete  section 
on  the  compressive  side  of 
the  neutral  plane  and  pc  the 
maximum  compressive  stress  in 
it,  we  have,  assuming  the  com- 
pressive stress  to  vary  directly  as  the  distance  from  the 
neutral  plane, 

Aps  =  i  pjbx.  (1) 

We  know  the  position  of  the  steel  reinforcement,  and  calling 
its  distance  from  the  top  of  the  beam  h,  we  have  the  sum  of 
the  moments  of  the  stresses  about  the  neutral  axis  equal  to 
the  bending  moment,  or 


t 

I  * 

n  Q  n 

Fig.  69. 


Aps  (h  -  X)  +  i  pJ)X  .$x  = 


(2) 


Now  as  the  bending  is  considered  uniform  the  extension  of 
the  steel  on  the  tension  side  must  equal  the  contraction  of 
that  part  of  the  concrete  on  the  compression  side  which  is  at 
the  same  distance  from  the  neutral  axis  as  the  steel,  therefore, 
by  Hook's  law  (strains  vary  as  distance  from  neutral  plane) 


P 


Es  (h  -  x) 


(3) 


We  now  have  three  equations  from  which  we  can  find  the 
three  unknown  quantities  ps,  pc  and  x',  ps  and  pc  being  the 
total  stresses.  If  we  have  a  reinforced  concrete  column  under 
a  load,  the  parts  of  the  load  borne  by  the  concrete  and  steel 


MISCELLANEOUS  PROBLEMS.  145 

can  be  found,  because  the  change  of  length  of  the  two  mate- 
rials must  be  the  same.     We  have,  therefore,  the  equations 

IPs          lpc   . 


ASES      ACEC 
and  the  load 

L  =  ps  +  Pc,  (2) 

where  I  is  the  length  of  the  column,  As  and  Ac  the  sectional 
areas  of  steel  and  concrete  respectively.  Concrete  columns 
are  usually  made  so  short  that  bending  does  not  enter  into 
the  calculations,  which  is. the  case  when  the  column  is  not 
longer  than  twelve  times  its  least  diameter.  A  difficulty  in 
investigating  the  bending  of  reinforced  concrete  beams  lies 
in  the  fact  that  the  modulus  of  elasticity  of  concrete  under 
compression  decreases  somewhat  as  the  loads  increase.  This 
will  evidently  cause  an  upward  movement  of  the  neutral 
plane.  Several  theories  with  different  assumptions  have  been 
advanced,  but  the  investigation  of  these  beams  may  be  con- 
sidered to  be  still  in  an  experimental  stage.  The  use  of  the 
equivalent  section  gives  fair  results.  Another  theory  assumes 
the  stress  in  the  concrete  to  vary  as  the  ordinates  of  a  pa- 
rabola and  that  the  moduli  of  elasticity  for  the  different 
stresses  are  those  found  by  tests  for  varying  pressures. 

79.  Poisson's  Ratio.  —  Experiment  shows  us  that  when  a 
piece  of  material  is  stretched,  the  area  of  a  transverse  section 
is  reduced  somewhat,  in  addition  to  the  change  made  in  the 
length  of  the  piece.  This  change  of  cross-sectional  area  is 
quite  marked  if  we  stretch  a  piece  of  rubber  because  the  elon- 
gation is  great ;  but  with  materials  such  as  are  used  in  build- 
ing the  change  of  cross-section  is  scarcely  perceptible,  for  the 
elongation  is  very  slight.  Tests  show  that  the  lateral  expan- 
sion or  contraction,  due  to  compression  or  tension,  bears  a 
constant  ratio  to  the  change  of  length.  This  is  known  as 
Poisson's  ratio.  If  then  the  extension  per  unit  length,  of  a 
round  rod  under  tension,  is  e}  the  length  of  its  diameter,  d, 


146 


STRENGTH  OF   MATERIAL. 


will  be  reduced  an  amount  equal  to  ked,  where  k  is  the  value 
of  Poisson's  ratio.  Poisson's  ratio  is  for 

Steel 297 

Iron 277 

Copper 340 

Brass 357 

If  a  rod  of  length  Z,  and  of  unit  square  section,  is  under  a 
tensile  stress,  F,  along  its  axis  and  a  compressive  stress,  R,  per- 
pendicular to  the  axis  on  two  opposite  faces,  the  total  exten- 
sion in  the  direction  of  the  axis,  is  that  due  to  the  tensile 
stress  plus  that  due  to  the  compression  stress,  or, 

Fl      kRl 

total  extension  =  —  H • 

E        E 

80,    Stress  in  Guns,  —  We   found  in  Art.  10  that  the  hoop 
stress  of  a  thin  cylinder  under  the  internal  pressure  of  S  Ibs. 

97* 

per  unit  area  was,  p  =  —  ,  where  t  was  the  thickness  of  the 

metal  and  r  the  radius  of  the  cylinder.      In  this  case  the 

metal  was  considered  so  thin, 
that  the  stress  might  be  taken 
as  uniform  throughout  the  sec- 
tion. If,  however,  the  metal 
is  thick,  we  cannot  use  this 
formula,  for  the  stress  will 
not  be  uniform  throughout  the 
section,  but  will  vary  in  some 
way  with  the  distance  from 
the  axis  of  the  cylinder. 

Let    us    consider    a    closed 
Flg'     '  cylinder  of  internal  radius  r2 

and  external  radius  rl  under  an  internal  pressure  S  and  an  ex- 
ternal pressure  $t  per  unit  area.  The  cylinder  having  closed 
ends  there  will  obviously  be  a  longitudinal  stress,  which  is 


MISCELLANEOUS   PROBLEMS.  147 

readily  seen  to  be  the  total  pressure  on  an  end  divided  by 
the  sectional  area  of  the  cylinder.  We  will  call  this  longi- 
tudinal stress  T  and  will  assume  it  is  uniform  throughout 
the  section.  If  now  we  consider  the  forces  acting  on  any 
element  of  volume  within  the  material  of  the  cylinder  the 
element  will  be  in  equilibrium  under  the  action  of  the  longi- 
tudinal stress  T,  the  hoop  stress  H  and  the  radial  stress  R. 
From  the  preceding  article  we  have  for  a  value  of  the  total 
extension  at  the  element  in  the  direction  of  the  axis  of  the 
cylinder 

e  =  -  (T  -  kH  -  kR), 
E 

k  being  the  value  of  Poisson's  ratio  for  the  material.  Now, 
if  sections  remain  plane  while  stressed,  we  may  assume  the 
total  extension  due  to  these  three  forces  constant,  and  T 
being  constant,  we  must  have  H  plus  R  equal  to  a  constant, 

H  +  R=C.  (1) 

Let  Fig.  70  represent  a  right  section  of  the  cylinder  of  unit 
thickness  (perpendicular  to  the  paper)  and  let  the  circular 
element  shown  be  of  width  dr.  If  we  call  the  radial  stress 
at  the  inner  surface  of  the  element,  R,  the  hoop  stress,  at  the 
inner  surface  of  the  element,  will  be  (Art.  10)  equal  to  Rr. 
The  radial  stress,  at  the  outer  surface  of  the  element,  will  be 
R  +  dR,  and,  as  the  radius  here  is  r  +  dr,  the  hoop  stress,  at 
the  outer  surface  of  the  element,  will  be  (R  -f  dR)(r  +  dr). 
The  hoop  stress  over  the  sectional  area  of  the  element  will 
be  (R  +  dR)  (r  +  dr)  -  Rr.  But  we  have  called  the  hoop 
stress  at  any  point,  H',  therefore  the  hoop  stress  on  this 
element  will  be  H,  multiplied  by  the  sectional  area  of  the 
element;  hence  we  will  have,  the  element  being  of  unit 
thickness, 


(R 

or,  neglecting  the  higher  order  of  infinitesimals, 

rdR  +  Rdr  =  Hdr.  (2) 


148  STRENGTH   OF   MATERIAL. 

From  (1)  we  have  H  =  C  -  R,  and,  substituting, 

rdR  +  Rdr=  Cdr  -  Rdr; 
multiplying  both  sides  of  the  equation  by  r,  we  have 

r2dR  +  2  rRdr  =  Crdr, 
or, 

d  (r*R)  -  Crdr, 
and,  integrating, 


where  Cv  is  the  constant  of  integration. 
From  the  last  equation  we  get 


and  substituting  this  value  of  R  in  (1)  we  get 

.-£-> 

Now  the  cylinder  has  an  internal  pressure  of  S  Ibs.  per  unit 
area  and  an  external  pressure  of  Sl  Ibs.  per  unit  area,  so  if 
r  —  r2  we  have  R  at  the  inside  surface  equal  to  S,  and  if 
r  =  rl  we  have  R  at  the  outside  surface  equal  to  Siy  and,  sub- 
stituting, we  get 


and 

SI=-  +  -T;  (6) 

2       TV*' 
solving  (5)  and  (6)  we  have 

Ci_r.VCS.-fl)  (7) 

7*o  7*i 


MISCELLANEOUS   PROBLEMS.  149 

and 

C=  2  ^f  '  ""  ^S}  ,  (8) 

1  2 

substituting  the  values  from  (7)  and  (8)  in  (3)  and  (4)  we 
get 

R  =  r*Sl-rSS      rW  (£,-£) 

and 

r.'S,  -  r,'S   ,  r.V,'  (S,  -  S) 


.'-' 


r-r 


If  there  is  no  external  pressure,  as  in  the  cases  of  cast  guns 
and  hydraulic  pipes,  Sl  will  be  equal  to  zero  and  our  for- 
mulas become 


and 


From  equations  (11)  and  (12)  we  see  that  the  greatest  stress 
is  at  the  inner  surface  of  the  cylinder,  and  here  the  total 
extension  found  by  the  method  of  Art.  79  must  not  exceed 
the  elastic  limit  of  the  material.  It  will  be  noticed  that  the 
values  of  H  and  R  found  above  do  not  take  into  considera- 
tion the  lateral  contraction  of  the  material  mentioned  in  the 
preceding  article.  If  we  consider  this  lateral  contraction, 
and  call  Hl  the  stress  which  would  produce  an  extension 
equal  to  the  total  elongation  found  by  the  methods  of  Art. 
79,  we  would  have 

H,  =  H  -  kR  -  kT. 

For  the  material  used  in  the  construction  of  the  modern 
naval  gun,  the  value  of  Poisson's  ratio  is  J.  Putting  in  this 


150  STRENGTH  OF  MATERIAL. 

value  for  k  and  also  the  values  of  H  and  R  from  equations  (9) 
and  (10),  the  value  of  T  being  found  directly  from  the  ex- 
ternal and  internal  end  pressures  and  the  sectional  area  of 
the  cylinder,  to  be 


T  = 


we  have,  after  reducing,  for  the  true  hoop  stress 
r,'St-r,'S   ,  4r,V(S,-S) 


which  is  the  formula  used  for  the  hoop  stress  for  naval  guns. 
81.  Built-up  Guns.  —  The  guns  at  present  used  in  the 
navy  are  known  as  "  built-up  "  guns;  that  is,  they  are  built 
of  several  pieces  which  are  made  separately.  The  inner 
piece  is  called  the  tube,  outside  of  it  is  the  jacket,  and  outside 
of  the  jacket  are  the  hoops.  The  exterior  diameter  of  the 
tube  is  made  a  little  greater  than  the  interior  diameter  of 
the  jacket,  the  exterior  diameter  of  the  jacket  a  little  greater 
than  the  interior  diameter  of  the  first  hoop,  etc.  The  pro- 
cess known  as  assembling  consists  in  heating  the  jacket 
until  it  expands  sufficiently  just  to  slip  over  the  tube,  after 
which  it  is  allowed  to  cool.  In  cooling  the  jacket  contracts 
and  grips  the  tube  firmly,  putting  considerable  external 
pressure  on  it.  The  hoops  are  then  put  over  the  jacket  in 
the  same  way.  This  way  of  securing  the  several  parts 
together  is  called  the  method  of  hoop  shrinkage,  and  obvi- 
ously if  two  parts  are  assembled  by  this  method,  the  inner 
one  will  be  under  hoop  compression  and  the  outer  one 
under  hoop  tension.  Considering  a  cylinder  formed  of  two 
parts,  let  the  radii  after  assembling  be  rz,  r2  and  rlt  the  least 
radius  being  r3.  Let  R3,  R2  and  Rl  be  the  radial  stresses  at 
the  inner,  intermediate  and  outer  surfaces.  We  will  call 
the  difference  between  the  inside  radius  of  the  jacket  and 
the  outside  radius  of  the  tube  before  assembling,  e:then  if 


MISCELLANEOUS   PROBLEMS.  151 

after  assembling  e2  is  the  decrease  in  the  length  of  the  radius 
of  the  tube  and  el  the  increase  in  the  length  of  the  radius 
of  the  jacket,  we  have 

e  =  e2  +  er  (1) 

If  H2  is  the  hoop  compression  on  the  outside  surface  of  the 
tube  and  Hl  the  hoop  tension  on  the  inside  surface  of  the 
jacket,  e2,  the  change  of  length  of  the  radius  of  the  tube  due 

to  the  stress  H2,  will  be  — —  (Chapter  I),  and  e^  the  change 

E 

of  length  of  the  radius  of  the  jacket  due  to  the  stress  Hlt  will 
be ;  hence 


E  E 

or, 

#2  +  ^=-'  (2) 

There  being  no  internal  pressure  in  this  cylinder,  equation 
(13)  of  the  preceding  article  will  give  us  a  formula  for  the 
hoop  compression  at  the  outer  surface  of  the  tube,  if  we  let 
S  =  0,  Si  =  R2  and  r  =  r2,  at  the  same  time  changing  r2  of 
equation  (13)  to  r3  and  rx  to  r2.  This  gives  us  the  equation 


2  o  r  2    ,-  2 

r2    --  r3         6  r2    (r2    •-  r3  ) 

There  being  no  external  pressure  on  the  jacket,  equation  (13) 
of  the  preceding  article  will  also  give  us  a  formula  for  the 
hoop  tension  at  the  inner  surface  of  the  jacket  if  we  put 
Sl  =  0,  S  =  R2  and  r  =  r2.  This  gives  us  the  equation 


152  STRENGTH  OF   MATERIAL. 

Taking  the  factor  R2  from  the  right  side  of  equations  (3) 
and  (4)  all  the  quantities  that  remain  are  known  (they  being 
the  different  radii).  Calling  the  value  of  these  remaining 
quantities  in  equation  (3)  Ci  and  those  for  equation  (4)  C2, 
we  have 

#2=^1  (5) 

and 

#i  =  -ttA;  (6) 

dividing  (5)  by  (6)  to  eliminate  R2  we  have 

!--!• 

Solving  the  simultaneous  equations  (2)  and  (7)  for  Hl  and 
#2  gives 

CEe 


and 


Equations  (8)  and  (9)  give  the  hoop  stresses  at  the  joint, 
and  having  these  we  can  from  either  of  the  equations  (3)  or 
(4)  find  the  value  of  R2.  Having  R2,  equation  (13)  of  the 
preceding  article  will  give  us  the  value  of  the  hoop  com- 
pression at  the  inner  surface  of  the  tube,  at  which  place  the 
hoop  compressive  stress  is  obviously  greatest  when  the  gun 
is  at  rest.  At  the  instant  of  firing  a  gun  the  internal  pres- 
sure becomes  very  great  and  both  the  tube  and  jacket  will 
then  be  under  hoop  tensile  stress,  but  obviously  the  first 
effect  of  the  internal  pressure  must  be  to  overcome  the 
initial  compressive  stress  in  the  tube,  though  it  increases 
the  tensile  stress  in  the  outer  hoops  where  the  effect  of  the 
internal  pressure  is  least;  thus  the  built-up  gun  can  sustain 
greater  internal  pressures  than  the  solid  gun. 


MISCELLANEOUS  PROBLEMS.  153 

82.  Stress  Due  to  a  Centrifugal  Force.  —  If  a  body  of  mass 
m,  at  the  end  of  a  cord,  is  whirled  round  in  a  circle  the 
cord  will  be  put  in  tension,  the  centrifugal  force,  as  we  have 

learned  from  mechanics,  being  equal  to ,  where  v  is  the 

linear  velocity  and  r  the  distance  of  the  center  of  gravity  of 
the  mass  from  the  axis  of  rotation.  If  n  is  the  number  of 
revolutions  per  second,  the  angular  velocity  aj  will  be  equal 
to  2  nn.  As  the  linear  velocity  is  raj,  the  tension  of  the  cord 
in  terms  of  the  angular  velocity  will  be  equal  to 

4  WxWr  I         W 


/  \ 

[m  —  —  and  a>  =  2nn  1  • 

V          g  I 


Consider  a  fly-wheel,  the  spokes  of  which  represent  a  com- 
paratively small  part  of  the  total  weight,  so  that  we  may 
consider  all  the  weight  as  being  at  the  rim  of  the  wheel. 
Let  the  thickness  of  the  rim  be  t  and  its  distance  from  the 
axis  of  rotation  be  r.  Let  the  weight  of  the  material  be 
w  Ibs.  per  unit  volume,  and  the  breadth  of  the  rim  a  units. 
Then  the  total  weight  of  the  rim  will  be  w .  2  nrta.  The 
total  radial  force  from  the  above  formula  is 

w  .  2  nrta  X  4  xWr 


and  the  radial  force  per  unit  area  will  be 

_  w  .  2  xrta  X  4  x2n2r     w  .  4  n2n2rt 
9(2xra)  ~ 

This  will  give  us  the  radial  stress  for  a  rim  so  thin  that  the 
stress  may  be  considered  uniform  throughout  the  section. 
If,  however,  we  consider  a  thick  rim  whose  inside  radius  is 
r2  and  whose  outside  radius  is  r1?  the  increment  of  radial 
stress  on  a  circular  element  whose  thickness  is  dr  will  be 


154  STRENGTH  OF  MATERIAL. 

given  by  the  above  formula.     We  will  have  then  for  the 
increment  of  radial  stress  on  this  elemental  volume 


w  .  4  xWrdr 
dR  =  -  -,  (1) 


the  value  of  this  increment  having  the  negative  sign  be- 
cause the  radial  stress  decreases  as  the  distance  from  the 
axis  of  rotation  increases.  To  prove  this  latter  statement, 
suppose  a  bar  of  uniform  section  and  length  /  rotates  about 
an  axis  through  its  end.  Let  a  be  the  sectional  area,  w  the 
weight  per  unit  volume,  and  x  the  distance  of  any  point  P 
from  the  axis  of  rotation.  The  stress  at  any  point  P  is  due 
to  the  centrifugal  force  of  the  part  of  the  rod  beyond  P; 
therefore,  as  the  center  of  gravity  of  this  part  is  at  a  dis- 

l  +  x 
tance from  the  axis  and  its  weight  is  a  (I  —  x}w,   the 

£l 

centrifugal  force,  by  the  formula,  at  the  beginning  of  this 
article  is 

c.r.-TO(*2~*2)a'2. 


It  is  clear  from  this  equation  that  when  x  =  Z  the  C.  F.  is 
zero,  and  when  x  =  0  it  is  a  maximum.  Proceeding  then 
the  integration  of  (1)  gives 

4  7i2n2wr2 


Ct  being  the  constant  of  integration.  We  know  the  value 
of  the  radial  stress  at  the  outer  edge  of  the  rim  is  zero; 
therefore  when  r  =  r 


MISCELLANEOUS  PROBLEMS.  155 

from  which 


hence, 

R  =  ^  W  -  H).  (2) 

^  0 

This  equation  gives  the  radial  stress  at  any  distance  r  from 
the  axis  of  rotation. 

Now  considering  the  equilibrium  of  any  particle,  we  have, 
by  the  same  reasoning  employed  in  Art.  80,  the  same  two 
equations, 

H  +  R  =  C  (3) 

arid 

rdR  +  Rdr  =  Hdr.  (4) 

Substituting  in  (4)  the  values  of  R  and  dR  from  (1)  and 
(2)  we  have 

4  xWwrdr        4  n*n2w 

r  •  -  +  (r*  -  r2)  dr  =  Hdr. 

9  20 

Integrating,  the  constant  of  integration  being  zero  when 
r  =  0,  we  have 

H  -  ^  W  +  r>).  (5) 

^  0 

Equations  (2)  and  (5)  give  the  radial  and  hoop  tensile  stress 
respectively  for  any  point  at  a  distance  r  from  the  axis  of 
rotation,  and  both  are  independent  of  r2.  Therefore  these 
equations  can  be  applied  to  solid  wheels,  such  as  a  grindstone, 
as  well  as  to  fly-wheels.  As  in  the  preceding  article  these 
formulae  have  been  deduced  without  considering  the  lateral 
deformation  of  Art.  79.  The  true  hoop  tension,  taking  this 
lateral  deformation  into  consideration,  would  be  Hl=H  —  kR 


156  STRENGTH  OF  MATERIAL. 

and  the  true  radial  stress  Rt  =  R  —  kH.  The  value  of  the 
radial  stress  is  greatest  where  r  =  r2  (equation  (2)),  and  for 
this  value  of  r  we  have 

R,=  -(V  -  krS  -  r22  -  A;r22).  (6) 

5/ 

The  hoop  tension  is  greatest  where  r  =  rl  ;  hence 


From  equations  (6)  and  (7)  it  is  clear  that  the  hoop  tension, 
being  the  greater  stress,  will  be  the  cause  of  rupture  and  also 
that  the  rupture  will  begin  at  the  outside  surface  of  the  rim. 
83.  Bending  Due  to  Centrifugal  Force.  —  We  will  have  a 
bending  moment,  due  to  centrifugal  force,  if  both  ends  of  a 
straight-  rod  are  constrained  to  move  in  circles,  for  example 
the,  horizontal  rod  between  the  two  driving  wheels  of  a  loco- 
motive. In  this  case  each  point  of  the  rod  at  any  instant 
is  moving  in  a  circle,  and  the  centrifugal  force  due  to  this 
motion  acts  in  the  direction  of  the  center  of  that  circle  for 
that  instant.  The  stress  caused,  acts  in  a  diametrically  oppo- 
site direction.  Obviously  the  effect  will  be  greatest  when  the 
horizontal  rod  is  at  its  lowest  point,  for  here  the  weight  of  the 
rod  itself  acts  down,  as  does  also  the  stress  due  to  centrifugal 
force.  Calling  the  radius  of  the  circle  described  by  the  ends 
of  the  rod  r,  and  w  the  weight  per  unit  length  of  the  rod 
(considered  of  uniform  cross-section)  ,  the  radial  stress  due  to 
centrifugal  force  will  be 


gr 

The  value  of  the  linear  velocity  v  is  obtained  from  our 
knowledge  of  the  speed  of  the  engine,  radius  of  the  drive 
wheels,  etc.  For  example,  if  the  speed  of  the  engine  is  S, 


MISCELLANEOUS  PROBLEMS. 


157 


radius  of  the  drive  wheels  r1;  and  radius  of  motion  of  the 
ends  of  the  rod  r,  the  linear  velocity  of  any  point  in  the 


horizontal  rod  will  be  S 


R 


,  and  this  gives 
wS2r 


This  value  of  R  is  uniform  throughout  the  length  of  the  rod; 
therefore  the  rod  may  be  considered  as  a  beam  loaded  uni- 
formly with  R  per  unit  length,  and  will  have  in  addition 


Fig.  71. 

when  at  its  lowest  position  a  uniform  load  due  to  its  own 
weight.     The  stress  at  any  point  can  then  be  readily  found 

p       M 

from  the  formula  for  bending,  —  =  —  • 

y      l 

A  connecting  rod  offers  an  important  example  of  bending 
stress  due  to  centrifugal  force,  but  in  this  case  one  end  of  the 
rod  is  constrained  to  move  in  a  straight  line,  while  the  other 
end  moves  in  a  circle.  At  the  instant  the  connecting  rod  is 
at  right  angles  with  the  crank  arm  (obviously  the  greatest 
effect  is  produced  when  the  load  is  perpendicular  to  the  con- 
necting rod),  if  the  end  of  the  connecting  rod  is  above  the 
center  of  motion  of  the  crank  arm,  the  weight  of  the  con- 
necting rod  will  act  down,  while  the  line  of  action  of  the 


158  STRENGTH  OF  MATERIAL. 

stress  due  to  the  centrifugal  force,  will  act  upward,  in  a  direc- 
tion parallel  to  the  crank  arm.  Therefore,  as  before,  the 
greater  stress  will  be  produced  when  the  end  of  the  connect- 
ing rod  is  below  the  center  of  motion  of  the  crank  arm.  In 
this  position,  however,  the  force  acting  on  the  piston  will  put 
the  connecting  rod  in  tension,  so  that  it  can  readily  sustain 
this  bending  stress.  When  the  end  of  the  connecting  rod 
is  above  the  center  of  motion  of  the  crank  arm,  the  stress, 
produced  by  the  pressure  on  the  piston,  is  compressive  and 
the  connecting  rod  is  in  the  condition  of  a  strut;  any  bend- 
ing due  to  centrifugal  force  will  now  become  an  important 
matter  (Chapter  XI).  Taking  the  position  then  as  shown 
in  Fig.  71,  the  centrifugal  load  at  any  point  of  the  con- 
necting rod  will  vary  directly  as  the  distance  of  the  point 
from  A;  for  as  the  radial  stress  at  any  point  due  to  the 

wv2 

motion  of  the  connecting  rod  is  equal  to  —  »  where  r  is  the 

gr 

radius  of  the  circle  being  described  by  the  point  at  the  in- 
stant and  the  value  of  v  is  the  linear  velocity  of  the  point  B, 
because  the  whole  connecting  rod,  at  the  instant,  is  mov- 
ing in  a  direction  tangential  to  the  circle  described  by  B. 
Obviously  r  varies  from  a  at  B  to  an  infinite  length  at  A; 
therefore,  we  must  have  the  load  on  the  connecting  rod  due 
to  the  centrifugal  force  vary  uniformly  from  zero  at  A,  to  a 

wv2 

maximum  at  B,  where  it  is  equal  to  -   -  •     The    bending 

ga 

moment  due  to  this  load  (Art.  42),  taking  A  for  the  origin,  is 


wv2  x\ 

<  = (Ix ) 

6  ga\  I 


I  being  the  length  of  the  connecting  rod  and  x  the  distance 
along  it  from  A  to  any  point.  This  bending  moment  tends 
to  bend  the  rod  so  that  the  middle  of  it  will  move  upward. 
The  weight  of  the  rod  itself  acts  vertically  downward,  and  the 


MISCELLANEOUS  PROBLEMS.  159 

component  of  the  weight  which  acts  perpendicular  to  the  rod 
will  be  w  cos  0  per  unit  length.  This  load  causes  a  bending 
moment  equal  to  (Art.  38) 


the  effect  of  which  is  to  cause  the  middle  of  the  rod  to  move 
downward.  The  total  bending  moment  at  any  point  distant 
x  from  A  is  then 


Having  the  bending  moment  at  any  point  due  to  the  motion 
and  weight  of  the  rod,  the  stress  due  to  these  causes  is  readily 

found  from  —  =  —  •     But  it  may  be  repeated  that  a  con- 

necting rod,  in  addition  to  the  above  stress,  suffers  a  com- 
pressive  stress,  when  in  this  position,  due  to  the  pressure  on 
the  piston,  and  also  a  bending  stress  due  to  this  compressive 
load,  which  puts  the  rod  in  the  condition  of  the  column  or 
strut  discussed  in  Chapter  XI.  Referring  to  that  chapter,  it 
will  be  seen  that  the  comparatively  slight  bending,  due  to 
centrifugal  force,  assumes  important  dimensions  when  we 
consider  that  the  least  variation  of  the  axis  of  the  connecting 
rod,  from  the  line  of  action  of  this  large  compressive  load, 
makes  the  opportunity  for  the  load  to  cause  bending. 

84.  Flat  Plates.  —  Experiment  has  proved  that  a  circular 
flat  plate  when  subjected  to  too  great  a  pressure  on  one  side 
always  breaks  along  a  diameter.  Any  diameter,  then,  of  a 
circular  plate,  is  perpendicular  to  the  greatest  tensile  stress 
due  to  bending,  caused  by  the  pressure  on  one  side  of  the  plate. 
Let  us  consider  a  circular  plate  simply  supported  at  the  cir- 
cumference, and  subjected  to  a  uniformly  distributed  load  on 
the  side  opposite  to  the  support.  If  the  plate  were  fixed  at 


160 


STRENGTH  OF   MATERIAL. 


the  circumference,  it  would  be  stronger  than  the  one  we  are 
considering,  just  as  a  beam  fixed  at  the  ends  is  capable  of 
supporting  a  greater  load  than  if  it  were  free  at  the  ends. 
We  will  consider  our  support  as  a  ring  on  which  the  plate 
rests,  the  pressure  being  uniformly  distributed  on  the  upper 
side. 

Let  Fig.  72  represent  such  a  plate,  the  support  being  at  the 
circumference,  and  the  load  w  Ibs.  per  unit  area.     We  must 


Fig.  72. 
first  find  the  bending  moment  at  the  diameter  AB.     The 

T 

area  of  the  triangular  element  shown  is   —  .  rdd,  and  the 


load  on  it  is  w  —  dd.     The  supporting  force  under  the  arc 
ft 

rdd  must  be  equal  to  this  load,  and  its  distance  from  the 
diameter  AB  is  r  cos  6.  The  center  of  gravity  of  the  uni- 
form load  on  the  triangular  element  is  at  a  distance  §  r  from 
the  center  and  §  r  cos  0  from  the  diameter  AB.  The  bend- 
ing moment  at  the  diameter  AB  due  to  these  two  forces 
is  then 


WT 

dM=  -- 


MISCELLANEOUS  PROBLEMS.  161 


Integrating  between  the  limits  -    -  and  —  for  6, 

2          2 


wr3 
M  = 


which  is  the  bending  moment  due  to  the  loads  on  one  side  of 
the  diameter  AB.  Letting  t  be  the  thickness  of  the  plate 
the  moment  of  inertia  of  the  section  through  AB  about  the 

rt3 
neutral  axis  is  /  =  - —  •      We  have  then  from  the  equation 

of  bending,  —  =  — ,   ly  in  this  equation  being  equal  to—  j 
y        I      \  2  / 


_wr3    t_     6  r> 

P  '  :~3~"~2'7?  ~  WT^ 

which  gives  the  stress  on  any  section  through  a  diameter.  In 
designing  a  plate  of  this  kind  we  would  know  the  pressure  to 
which  it  would  be  subjected  and  the  dimensions  of  the  open- 
ing it  would  cover,  so  if  we  put  the  limiting  tensile  strength 
of  the  material  we  are  to  use  for  p,  we  can  solve  the  above 
equation  for  t,  the  thickness  necessary. 

If  the  load  instead  of  being  uniform  is  a  concentrated  one, 
calling  it  W  and  supposing  it  is  to  be  at  the  most  effective 
position,  the  center,  and  to  bear  uniformly  on  a  small  circle 
of  radius  ^  concentric  with  the  supporting  ring,  the  part  of 

W 
the  load  on  one  side  of  any  diameter  will  be — ,and  its  center 

4  r 
of  gravity  will  be  at  a  distance  — !  from  the  diameter.    The 

3  7i 

supporting  force  will  be  uniform  throughout  the  length  of 
the  semicircle  of  the  ring  on  the  same  side  of  the  diameter  as 
this  load,  and  the  center  of  gravity  of  this  semicircle  is  at  a 


162  STRENGTH  OF  MATERIAL. 

2r 
distance  -  -  from  the  diameter,  the  total  supporting  force 

W 
on  this  semicircle  being  —  .      The  bending  moment  at  the 

diameter  will  be 


t  being  the  thickness  of  the  plate,  and  the  moment  of  inertia 

rts 
of  the  section  being  7  =  —  as  before,  we  have  for  the  stress  in 

the  section 

My      W  /         2  rA    t       6        3  W  /         2  r,\ 
p  =  -  -  =  — [r _I|.  _  .  _  1 1 M  . 

7          TT  V  3  /    2     rt*        xt2   \         3  r ) 

If  r,  =  r,  or,  which  is  the  same  thing,  if  the  plate  bears  a 
uniform  load,  we  have,  remembering  W  will  now  be  equal 
to  nr2w , 

r2 

/2 

and  if  rt  =  0,  or  the  load  is  concentrated  at  a  point, 

3  W 


The  stress  given  by  these  formulae  is  the  maximum  stress, 
which  occurs,  as  one  would  expect,  at  the  center  of  the  plate. 
Experiment  proves  this  to  be  the  case,  for  such  plates  begin 
to  crack  at  the  center  and  the  crack  extends  from  there,  along 
a  diameter,  to  the  edge  of  the  plate.  Experiment  also  proves 
the  above  formula  to  give  very  approximate  results,  so, 
though  this  may  be  a  tentative  method  of  deducing  them  (the 
assumptions  not  being  strictly  true),  they  will  serve  for  all 
practical  purposes. 


MISCELLANEOUS  PROBLEMS. 


163 


---a — 
Fig.  73. 


Rectangular  Plates.  —  Experiment  shows  that  rectan- 
gular plates,  when  the  length  is  not  more  than  about  twice 
the  width,  will  crack  along 
a  diagonal.  Therefore,  if  the 
sides  are  a  and  b  (Fig.  73), 
the  diagonal  section  will  sup- 
port the  greatest  bending 
stress.  Suppose  the  plate  of 
thickness  t  to  support  a  uni- 
form load  of  w  per  unit  area. 

wab 

The  load  on  one  side  of   the    diagonal  will  be and  its 

JL 

*  or 
center  of  gravity  will  be  at  a  distance  -  from  the  diagonal. 

o 

Assuming  the  supporting  force  to  be  uniform  along  the  edges, 
the  part  acting  along  the  edge  a  will  act  at  its  center  of 

gravity,  which  is  at  a  distance  -  from  the  diagonal,  as  is  also 

fj 

the  center  of  gravity  of  the  part  acting  along  the  edge  b. 
These  two  edges  will  support  the  whole  load  on  this  side  of 
the  diagonal,  or  the  whole  supporting  force  here  will  equal 

— —  •     The  bending  moment  about  the  diagonal  then  is 


wab  x       wab  x       wabx 

TM  .  . I      

23         2      2   "       12 


Calling  g  the  length  of  the  diagonal,  the  moment  of  inertia 

at3 

of  the  section  through  it  about  its   neutral  axis  is  /=  — 

1 2, 

and  y  =  - .     The  equation  of  bending  gives  us 

My      wabx    t    12       wabx 
12    *2  'at*  ~  2qt2  ' 


164  STRENGTH  OF  MATERIAL. 

Referring   to   the   figure,   g  =  \/a2  +  b2,  and   from   similar 

Da 
triangles  x  =  —   -         ;  substituting  these  values, 

Va2  +  b2 

wa2b2 
p  = 


2  (a2  +  b2)  t2 
If  the  plate  is  square,  a  =  b,  and  we  have 

wa2 

P  =  4?' 

These  formulae  should  for  practical  purposes  have  a  factor 
on  the  right  member  of  the  equation  of  1.5  if  the  support  is 
&  fixed  one  (riveted  joint),  and  of  5  if  there  is  only  a  simple 
support. 

The  formula  for  elliptical  plates,  which  experiment  shows 
break  along  the  major  axis,  is,  for  wrought  iron  and  steel  plates 
simply  supported  around  the  edges, 

wa2b2 
P~*'  t2  (a2  +  b2} 

(if  cast  iron  is  used  the  coefficient  is  3  instead  of  f ) ,  where  w 
is  the  load  per  unit  area,  t  the  thickness,  and  a  and  b  the 
semi-major  and  semi-minor  axes  respectively.  The  theoret- 
ical solution  for  elliptical  plates  is  very  difficult  because  it 
involves  elliptical  integrals  and  because  the  supporting  force 
is  not  uniform  around  the  edge,  as  indeed  is  also  the  case 
for  rectangular  plates,  but  the  variation  for  the  latter  is  not 
excessive  unless  the  plate  is  more  than  two  or  three  times 
as  long  as  it  is  wide.  In  fact  all  the  above  deductions  are 
approximations,  for  the  assumptions  made  are  not  strictly 
true.  The  formula?,  however,  agree  closely  with  the  results 
obtained  experimentally  and  may  be  assumed  correct  for 
all  practical  purposes. 

The  subject  of  flat  plates  is  probably  the  most  unsatis- 
factory one  in  the  study  of  strength  of  material,  and  practi- 
cal engineers  have  different  methods  for  each  form  of  plate. 


MISCELLANEOUS  PROBLEMS.  165 

Fault  may  be  found  with  the  assumptions  of  most  of  these 
methods,  though  they  all  give  approximate  results. 

85.  The  principle  of  least  work  may  be  stated  as  fol- 
lows: For  stable  equilibrium  the  stresses  in  any  structure 
must  have  such  values  that  the  potential  energy  of  the  sys- 
tem is  a  minimum.  The  stresses  of  course  must  be  within 
the  elastic  limit  of  the  material.  When  forces  act  upon 
bodies  which  conform  to  Hook's  law,  the  principle  of  least 
work  may  be  applied  to  determine  some  unknown  reaction. 
Perhaps  the  best-known  example  is  that  of  a  four-legged 
table  which  supports  an  unsymmetrically  placed  load.  We 
can  get  from  our  knowledge  of  statics  three  equations  to 
find  the  part  of  the  load  supported  by  each  leg,  and  the  solu- 
tion can  be  arrived  at  by  means  of  the  fourth  condition  fur- 
nished by  the  principle  of  least  work. 

Distribution  of  Stress We  have  all  along  assumed  that 

the  stress  is  uniform  throughout  a  section.  As  a  matter  of 
fact  this  is  not  the  case.  It  can  be  mathematically  proved 
that  the  shearing  stress  in  a  rod  of  square  section  varies  as 
the  ordinates  of  a  parabola,  being  zero  where  the  normal 
stress  due  to  bending  is  a  maximum,  and  a  maximum  where 
the  normal  stress  due  to  bending  is  zero  (at  the  neutral 
surface).  Again,  the  shear  parallel  to  the  neutral  axis  in 
a  rod  of  square  section  is  zero,  but  in  a  rod  of  circular  section 
it  has  a  finite  value. 

Velocity  of  Stress.  —  When  a  force  is  applied  to  a  piece 
of  material  the  stress  is  not  instantaneously  produced,  but 
moves  with  a  wave  motion  through  the  mass.  The  velocity 
•  of  this  motion  can  be  found,  and  it  is  shown  to  depend 
upon  the  stiffness  and  density  of  the  material.  The  velocity 
of  stress  should  be  taken  into  account  in  problems  involving 
impact  and  suddenly  applied  loads. 

Internal  Friction.  —  When  material  is  subjected  to  force 
and  deformation  occurs  the  molecules  of  the  material  move 
and  this  motion  is  resisted  by  internal  friction.  Heat  is 


166  STRENGTH  OF  MATERIAL. 

produced  and  for  the  time  between  the  application  of  the 
load  and  the  complete  rearrangement  of  the  molecules  the 
stresses  at  planes  through  the  material  are  changing;  for 
this  time  then  our  formulae  for  planes  of  maximum  stress 
are  not  correct. 

Fatigue  of  Materials.  —  Experiment  proves  that  material 
will  break  if  subjected  to  repeated  stress,  even  if  the 
stress  be  somewhat  below  the  ultimate  strength  of  the 
material.  Experiment  also  shows  that  the  greater  the  num- 
ber of  applications  the  less  becomes  the  value  of  the  stress 
necessary  to  cause  rupture.  For  example,  about  a  hun- 
dred thousand  applications  of  a  stress  of  49,000  Ibs.  to 
wrought  iron  will  cause  rupture,  but  if  500,000  applications 
were  made  the  stress  need  be  only  39,000  Ibs.  The  loss  of 
strength  due  to  repeated  stress  is  known  as  the  fatigue  of 
materials.  This  fatigue  is  more  marked  if  the  stress  alter- 
nates from  tension  to  compression  and  back  again. 

The  preceding  facts  have  been  mentioned  to  give  the 
student  the  knowledge  of  their  existence  so  that  if  inclined 
he  may  look  them  up  in  more  complete  works  on  the  sub- 
ject. The  time  limit  of  the  course  for  which  this  book  has 
been  written  precludes  the  possibility  of  entering  into  dis- 
cussion of  many  subjects  of  importance,  such  as  the  stress 
in  hooks,  links,  springs,  rollers,  foundations,  arches  and 
many  others. 

Miscellaneous  Examples: 

1.  A  reinforced  concrete  beam  is  48  ins.  wide,  54  ins. 
long  and  5  ins.  deep.  It  has  a  sectional  area  of  3.6  sq.  in. 
of  steel,  the  center  of  gravity  of  which  is  I  ins.  below  the  top 
of  the  beam  which  is  uniformly  loaded  with  2400  Ibs.  per 

CTjl  \ 

Use  —  =10.)     Find  the  position  of  the  neu- 
Ec          ) 

tral  surface,  and  the  stress  in  the  steel. 
Ans.     Neutral  surface  3.45  insk  below  top. 


MISCELLANEOUS  PROBLEMS.  167 

2.  A  reinforced  concrete  beam  is  60  ins.  long,  48  ins. 
wide  and  5  ins.  deep.     The  sectional  area  of  the  steel  is  2.4 
sq.  in.  and  its  center  of  gravity  is  f  ins.  below  the  top  of  the 
beam.     The  load  is  uniform  and  equals  3600  Ibs.  per  in. 
length.     Find  the  position  of  the  neutral  surface  and  the 

771 

stress  in  the  concrete  and  steel.     — -  =  10. 

^c 

Ans.     Neutral  surface  1.68  ins.  from  top. 

3.  If  the  allowable  unit  stress  in  concrete  be  500  Ibs.  per 
sq.  in.,  and  that  for  steel  be  10,000  Ibs.  per  sq.  in.,  what  per- 
centage of  steel  must  be  put  in  a  beam  20  ins.  wide  by  10 
ins.  deep  if  the  steel  is  at  a  distance  of  9  ins.  from  the  top? 

Ans.     0.75%. 

4.  If  the  allowable  unit  stresses  for  concrete  and  steel  are 
500  and  25,000  Ibs.  per  sq.  in.  respectively,  what  is  the 
resisting  moment  of  a  beam  7  ins.  wide  and  10  ins.  deep  if 
the  reinforcement  of  steel  is  1%  of  the  sectional  area  and 
placed  at  the  lowest  point  of  the  beam? 

Ans.     92,800  in.-lbs. 

5.  A  beam  is  8  ft.  long  and  1  ft.  wide.     The  concrete  is 
5  ins.  thick  and  the  steel  reinforcement  is  1  in.  from  the  bot- 
tom of  the  beam.     What  area  of  steel  section  will  be  neces- 
sary?   Using  J  in.  square  bars,  what  should  be  the  distance 
between  them  if  a  floor  is  made  in  this  way? 

6.  The  cross-section  of  a  steel  bar  is  16  sq.  in.     The  bar 
is  put  under  a  stress  of  27,000  Ibs.  per  sq.  in.     If  k  —  J, 
what  is  the  sectional  area  while  under  stress? 

Ans.     15.99  sq.  in. 

7.  A  steel  bar  is  2.5  in.  in  diameter  and  18.5  ft.  long. 
What  are  its  length  and  sectional  area  under  a  pull  of 
64,000  Ibs.? 


168  STRENGTH  OF  MATERIAL. 

8.  The  inside  radius  of  a  cylinder  is  6  ins.,  its  outside 
radius  is  1  ft.     The  internal  pressure  is  600  Ibs.  per  sq.  in., 
and  the  external  pressure  is  that  due  to  the  atmosphere. 
What  are  the  hoop  and  radial  stresses  at  the  outside  and 
inside  surfaces,  also  midway  between  these  surfaces? 

Ans.     Hoop  stress  inside,  960;  outside,  375  Ibs.  per  sq.  in. 

9.  A  solid  cylinder  is  under  a  uniform  external  pressure 
of  14,000  Ibs.  per  sq.  in.     Show  that  the  hoop  and  radial 
stresses  are  uniform  and  give  values. 

10.  A  gun  is  built  of  a  tube,  inside  radius  3  ins.,  outside 
radius  5  ins.,  and  a  jacket  2  ins.  thick.     Before  assembling, 
the  difference  between  the  outside  radius  of  the  tube  and 
the  inside  radius  of  the  jacket  was  .004  in.     What  are  the 
stresses  at  the  outside  and  inside  surfaces? 

Ans.     Hoop  compression,  14,400  Ibs.  per  sq.  in. 

11.  In  example  10  what  powder  pressure  when  firing  the 
gun  will  just  reverse  the  stress  at  the  inner  surface  of  the 
tube? 

12.  The  radii  of  a  gun  composed  of  tube  and  jacket  are 
r3  =  3.04  ins.,  r2  =  5.8  ins.,  and  ^  =  9.75  ins.      The  allow- 
able unit  stress  is  50,000  Ibs.  per  sq.  in.  for  both  tension  arid 
compression.     What  are  the  radii  before  assembling? 

Ans.  Radius  of  bore,  3.0451  ins.;  outside  radius  of  tube, 
5.805  ins.;  inner  radius  of  jacket,  5.7915  ins.;  outer  radius, 
9.7436  ins. 

13.  What  internal  powder  pressure  will  stress  the  gun  of 
example  12  to  just  50,000  Ibs.  per  sq.  in.  tension? 

Ans.     51,100  Ibs.  per  sq.  in. 

14.  What  is  the  greatest  tangential  stress  in  a  cast-iron 
fly-wheel  30  ft.  in  diameter,  rim  1  in.  thick  and  4  ins.  wide, 
when  it  is  making  60  revolutions  per  minute? 

Ans.     Very  roughly,  800  Ibs.  per  sq.  in. 


MISCELLANEOUS  PROBLEMS.  169 

15.  What  is  the  stress  in  a  cast-iron  fly-wheel  rim  having 
a  linear  velocity  of  1  mile  a  minute? 

Ans.     Roughly,  750  Ibs.  per  sq.  in. 

16.  What  diameter  should  a  fly-wheel  have  if  it  is  to 
make  100  revolutions  per  minute,  and  the  maximum  allow- 
able linear  velocity  is  6000  ft.  per  min.? 

Ans.     Roughly,  19  ft. 

17.  A  cast-iron  bar  9  ft.  long,  3  ins.  wide  and  2  ins.  thick 
revolves  about  an  axis  }  in.  in  diameter  passed  through  it 
at  a  distance  of  4|  ft.  from  one  end.     How  many  revolutions 
per  second  will  produce  rupture? 

18.  A  solid  steel  circular  saw  is  4  ft.  in  diameter,  and 
makes  2700  revolutions  per  minute.     What  is  the  stress  at 
the  circumference?     How  many  revolutions  per  minute  will 
cause  a  stress  of  35,000  Ibs.? 

19.  An  engine  is  making  750  revolutions  per  minute;  the 
connecting  rod  is  2  ft.  long,  and  the  crank  arm  6  ins.,  ma- 
terial steel.     What  is  the  bending  stress,  due  to  centrifugal 
force,  on  the  connecting  rod,  if  the  area  of  its  section  be  1.5 
sq.  ins.? 

20.  A  cast-iron  fly-wheel,  mean  diameter  of  rim  20  ft., 
makes  90  revolutions  per  minute.     The  cross-section  of  the 
rim  is  10  sq.  ins.     What  is  the  stress? 

21.  What  must  be  the  thickness  of  a  cast-iron  cylinder 
head  36  ins.  in  diameter,  allowable  stress  3600  Ibs.  per  sq.  in., 
to  sustain  a  load  of  250  Ibs.  per  sq.  in.? 

Ans.     5  ins. 

22.  The  allowable  stress  for  steel  being  12,000  Ibs.  per  sq. 
in.,  how  thick  would  a  steel  head  for  the  cylinder  of  example 
21  have  to  be? 

Ans.     2.6  ins. 


170  STRENGTH  OF  MATERIAL. 

23.  A  circular  steel  plate  is  24  ins.  in  diameter  and  1.5 
ins.  thick.     It  carries  a  load  of  4000  Ibs.  at  the  center  rest- 
ing on  a  circle  of  1  in.  diameter.     What  is  the  maximum 
stress? 

Ans.     1650  Ibs.  per  sq.  in. 

24.  Suppose  the  load  of  example  23  were  distributed  on  a 
surface  of  3  ins.  diameter.     What  would  be  the  stress? 

Ans.     1555  Ibs.  per  sq.  in. 

25.  What  must  be  the  thickness  of  a  steel  plate  4  ft. 
square  to  carry  200  Ibs.  per  sq.  ft. 

Ans.     0.2  in. 

26.  What  uniform  load  can  be  carried  by  a  wrought-iron 
plate  f  in.  thick,  5  ft.  long  and  3  ft.  wide? 

Ans.     About  12  Ibs.  per  sq.  in. 

27.  A  floor  18  ft.  long,  15  ft.  wide  is  made  of  concrete 
4  ins.  thick,  with  1  in.  square  wrought-iron  rods,  spaced  1  ft. 
apart  and  at  .75  in.  from  the  bottom  of  the  concrete.     The 
floor  carries  a  load  of  150  Ibs.  per  sq.  ft.     What  is  the  maxi- 

Es 

mum  stress?       —  =  15. 
Ec 

Ans.   Stress  is. about  450  Ibs.  per  sq.  in. 

28.  An  elliptical  plate  (cast  iron)  has  a  major  axis  24  ins. 
long,  a  minor  axis  16  ins.,  and  is  under  a  uniform  pressure  of 
22  Ibs.  per  sq.  in.     The  allowable  stress  is  3000  Ibs.  per 
sq.  in.,  and  the  plate  is  simply  supported  at  the  edges. 
What  must  be  the  thickness? 

Ans.     1  in. 


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